Question 2:<\/h4>\n\n\n\n
Can you ever have a situation in which a light ray goes undeviated through a prism?<\/p>\n\n\n\n
Answer:<\/h4>\n\n\n\n
There is only one such situation. Light rays go undeviated through a prism only when the prism is kept in a medium that has the same refractive index as that of the prism itself. In all other situations, there will always be some minimum deviation in the path of a light ray passing through the prism.<\/p>\n\n\n\n
Question 3:<\/h4>\n\n\n\n
Why does a diamond shine more than a glass piece cut to the same shape?<\/p>\n\n\n\n
Answer:<\/h4>\n\n\n\n
A diamond shines more because of the phenomenon of total internal reflection (TIR). The refractive index of diamond is \u22482.4 and that of glass is \u22481.5.
By using the relation<\/p>\n\n\n\n
\u03bc=1sin C, where C<\/em> is the critical angle, we get refractive index (\u03bc<\/em>). A narrow beam of light passes through a slab obliquely and is then received by an eye (figure 18-Q1). The index of refraction of the material in the slab fluctuates slowly with time. How will it appear to the eye? The twinkling of stars has a similar explanation. The beam of light will seem to appear and disappear, just like the twinkling of the stars. This is because the refractive index of the material in the slab fluctuates slowly with time. So, the light rays get refracted differently with time and this causes them to get concentrated during certain times, or get diffused at other times.<\/p>\n\n\n\n Can a plane mirror ever form a real image?<\/p>\n\n\n\n No, a plane mirror can never form a real image. This is because in all possible situations of image formation, the light rays never actually meet after getting reflected, but they only appear to meet behind the mirror, forming a virtual image always.<\/p>\n\n\n\n If a piece of paper is placed at the position of a virtual image of a strong light source, will the paper burn after sufficient time? What happens if the image is real? What happens if the image is real but the source is virtual?<\/p>\n\n\n\n No, the paper will not burn even after sufficient time. This is because the piece of paper is placed at the position of a virtual image of a strong light source, so light rays are not actually converging at that point, but seem to converge. As no light rays are concentrating on the point, the paper will never burn. Can a virtual image be photographed by a camera?<\/p>\n\n\n\n Yes, a virtual image can be photographed by a camera. This is because the light rays emitting from a virtual image and reaching the lens of the camera are real. A virtual image is formed from a reflecting surface after reflection of real incident rays, so these rays enter the camera to provide effects on the photographic film. Just like a virtual image can be seen by us in a mirror with our eyes, it can be photographed by a camera.<\/p>\n\n\n\n In motor vehicles, a convex mirror is attached near the driver\u2019s seat to give him the view of the traffic behind. What is the special function of this convex mirror which a plane mirror can not do?<\/p>\n\n\n\n The special function of a convex mirror is that it creates the image of a distant object that is reduced in size, is upright or erect and always lies within the virtual focal length of the mirror. A plane mirror cannot do this. Also, as the image is formed within the focal length, the image is close to the mirror as well as is small in size, enabling the driver to clearly view the nearer vehicles behind the motor vehicle.<\/p>\n\n\n\n If an object far away from a convex mirror moves towards the mirror, the image also moves. Does it move faster, slower or at the same speed as compared to the object?<\/p>\n\n\n\n The image of the object moves slower compared to the object. It can be explained using the mirror formula :<\/p>\n\n\n\n 1u+1v=1fWe know that for a convex mirror, the object distance (u<\/em>) is positive, image distance (v<\/em>) is negative and the focal length (f<\/em>) is also negative. Thus mirror formula of a convex mirror is:<\/p>\n\n\n\n 1u-1v=-1fAs u<\/em> = +ve<\/p>\n\n\n\n 1v-1f>01v>1fv<fTherefore, the image is always formed within the focal length of the mirror. Thus, the distance moved by the image is much slower than the distance moved by the object.<\/p>\n\n\n\n Suppose you are inside the water in a swimming pool near an edge. A friends is standing on the edge. Do you find your friend taller or shorter than his usual height?<\/p>\n\n\n\n When viewed from the water, the friend will seem taller than his usual height. \u03bc1-u+\u03bc2v=\u03bc2-\u03bc1RWhere refractive index of water is \u03bc<\/em>2<\/sub> and refractive index of air is \u03bc<\/em>1<\/sub>. \u03bc1-u+\u03bc2v=\u03bc2-\u03bc1\u221e\u03bc1u=\u03bc2vv=\u03bc2 \u03bc1\u00d7uAs \u03bc<\/em>1<\/sub> = 1 m=vuPutting the value of v<\/em> in the above equation:<\/p>\n\n\n\n m=u\u00d7\u03bc2um=\u03bc2As the magnification is greater than 1, so the apparent height seems to be greater than actual height.<\/p>\n\n\n\n The equation of refraction at a spherical surface is<\/p>\n\n\n\n \u03bc2\u03bd-\u03bc1\u03bc=\u03bc2-\u03bc1R.<\/p>\n\n\n\n Taking<\/p>\n\n\n\n R=\u221e, show that this equation leads to the equation<\/p>\n\n\n\n Real depthApparent depth=\u03bc2\u03bc1for refraction at a plane surface.<\/p>\n\n\n\n Proof:<\/p>\n\n\n\n \u03bc2v-\u03bc1u = \u03bc2-\u03bc1RNow R=\u221e \u03bc2v-\u03bc1u = \u03bc2-\u03bc1\u221e\u03bc2v-\u03bc1u =0 \u03bc2v=\u03bc1u\u03bc1\u03bc2 = uvBut uv=Real depth\/heightApparent depth\/height\u2234 Real depth\/heightApparent depth\/height=\u03bc1\u03bc2<\/p>\n\n\n\n \u03bcA thin converging lens is formed with one surface convex and the other plane. Does the position of image depend on whether the convex surface or the plane surface faces the object?<\/p>\n\n\n\n Yes. Using the lens maker formula we can show it. We know that the formula is:<\/p>\n\n\n\n 1f=\u03bc2\u03bc1-11R1-1R2Where, f<\/em> is the focal length of the thin converging lens. Case 1 1f=\u03bc2\u03bc1-11R1-1\u221e1f=\u03bc2\u03bc1-11R1 \u2026(i)Case 2<\/p>\n\n\n\n When the plane surface is facing the object, we have:<\/p>\n\n\n\n 1f=\u03bc2\u03bc1-11\u221e-1R11f=-\u03bc2\u03bc1-11R1 \u2026(ii)For Case 1, the focal length is positive and for the Case 2 the focal length is negative. Thus, the image distance is different in both cases.<\/p>\n\n\n\n A single lens is mounted in a tube. A parallel beam enters the tube and emerges out of the tube as a divergent beam. Can you say with certainty that there is a diverging lens in the tube?<\/p>\n\n\n\n Yes, we can say so with certainty, as only a diverging lens can cause a parallel beam of light to diverge.<\/p>\n\n\n\n An air bubble is formed inside water. Does it act as a converging lens or a diverging lens?<\/p>\n\n\n\n It will act as a diverging lens. This is because the refractive index of air is less than that of water. Thus, light will diverge after passing through the bubble.<\/p>\n\n\n\n Two converging lenses of unequal focal lengths can be used to reduce the aperture of a parallel beam of light without loosing the energy of the light. This increase the intensity. Describe how the converging lenses should be placed to do this.<\/p>\n\n\n\n Let the two converging lenses be L1<\/sub> and L2<\/sub>, with focal lengths f<\/em>1<\/sub> and f<\/em>2<\/sub> respectively. If a spherical mirror is dipped in water, does its focal length change?<\/p>\n\n\n\n No, the focal length of mirror will not change. This is because the focal length of a mirror does not depend on the refractive index of the medium in which the light rays travel.<\/p>\n\n\n\n If a thin lens is dipped in water, does its focal length change?<\/p>\n\n\n\n Yes, the focal length of the lens will change. This is because its focal length is dependent on the medium in which the light rays travel.<\/p>\n\n\n\n Can mirrors give rise to chromatic aberration?<\/p>\n\n\n\n No, mirrors cannot give rise to chromatic aberration. This is because chromatic aberration occurs due to the refraction of different colours of light. In case of mirrors, refraction of light does not take place.<\/p>\n\n\n\n A laser light is focussed by a converging lens. Will there be a significant chromatic aberration?<\/p>\n\n\n\n No, there will not be a significant chromatic aberration because a laser light is monochromatic in nature. Therefore, all light will get converged at the same point.<\/p>\n\n\n\n A point source of light is placed in front of a plane mirror. (a) All the reflected rays meet at a point when produced backward.<\/p>\n\n\n\n Here, the angle of reflection is equal to the angle of incidence. Therefore, all rays get reflected to converge at a single point to form the point image of the point source.<\/p>\n\n\n\n Total internal reflection can take place only if (b) light goes from optically denser medium to rarer medium<\/p>\n\n\n\n In this case the incident angle is greater than critical angle, so the light gets reflected back into the same denser medium, instead of being refracted .<\/p>\n\n\n\n In image formation from spherical mirrors, only paraxial rays are considered because they (c) form nearly a point image of a point source<\/p>\n\n\n\n Since, when reflected back, they meet at a single point forming a point image of a point source.<\/p>\n\n\n\n A point object is placed at a distance of 30 cm from a convex mirror of focal length 30 cm. The image will form at (d) 15 cm behind the mirror<\/p>\n\n\n\n Since u<\/em> = \u2212 30 cm From the mirror formula:<\/p>\n\n\n\n 1u+1v=1for1v=1f-1u\u21d21v=130-1-301v=130+1301v=230\u2234v=15 cm<\/p>\n\n\n\n Figure (18-Q2) shows two rays A<\/em> and B<\/em> being reflected by a mirror and going as A<\/em>\u2018 and B<\/em>\u2018. The mirror (a) is plane<\/p>\n\n\n\n This is because the reflected rays are still parallel, which is only possible if the mirror is a plane mirror. A spherical mirror will either converge or diverge the reflected rays.<\/p>\n\n\n\n The image formed by a concave mirror (c) is certainly real if the object is virtual<\/p>\n\n\n\n As a virtual object is the virtual image of the object, it is always formed beyond the focus of the concave mirror. Thus, the image formed by a concave mirror of the virtual object will always be real.<\/p>\n\n\n\n Figure (18-Q3) shows three transparent media of refractive indices<\/p>\n\n\n\n \u03bc1, \u03bc2 and \u03bc3. A point object O<\/em> is placed in the medium<\/p>\n\n\n\n \u03bc2. If the entire medium on the right of the spherical surface has refractive index<\/p>\n\n\n\n \u03bc1, the image forms at O<\/em>\u2018. If this entire medium has refractive index<\/p>\n\n\n\n \u03bc3, the image forms at O<\/em>\u201c. In the situation shown, (d) two images form, one at O<\/em>\u2018 and the other at O<\/em>\u201d<\/p>\n\n\n\n Light will be refracted differently in both mediums on the right side. Thus, two images will be formed, one at O<\/em>\u2018 due to refraction from medium \u03bc<\/em>1<\/sub> and another at O<\/em>\u201d due to refraction from medium \u03bc<\/em>3<\/sub>.\u03bc<\/em>1<\/p>\n\n\n\n Four modifications are suggested in the lens formula to include the effect of the thickness t<\/em> of the lens. Which one is likely to be correct? 1\u03bd-1u=tuf(b)<\/p>\n\n\n\n t\u03bd2-1u=1f(c)<\/p>\n\n\n\n 1\u03bd-t-1u+t=1f(d)<\/p>\n\n\n\n 1\u03bd-1u+tuv=tf.<\/p>\n\n\n\n (c)<\/p>\n\n\n\n 1v-t-1u+t=1fThe thickness of lens will cause a shift in the position of object and image from the ideal condition, which will be:<\/p>\n\n\n\n unew=u +t(1 + 1\u03bc ) and vnew=v \u2013 t(1 + 1\u03bc )or unew=u + t + t\u03bc and vnew=v \u2013 t + t\u03bcAs<\/p>\n\n\n\n t\u03bcwill be quite small, it can be ignored. 1v-t-1u+t=1f<\/p>\n\n\n\n A double convex lens has two surfaces of equal radii R<\/em> and refractive index<\/p>\n\n\n\n m=1\u00b75. We have, f=R\/2(b)<\/p>\n\n\n\n f=R(c)<\/p>\n\n\n\n f=-R(d)<\/p>\n\n\n\n f=2R.<\/p>\n\n\n\n (b) f<\/em> = R<\/em><\/p>\n\n\n\n As from lens maker formula:<\/p>\n\n\n\n 1f=(m \u2013 1)(1R1-1R2)\u21d21f=(1.5 \u2013 1)(1R-1-R) (As given R1=R2=R) 1f=(0.5)(1R+1R) 1f=(0.5)(2R) f = R=R<\/em><\/p>\n\n\n\n A point source of light is placed at a distance of 2 f<\/em> from a converging lens of focal length f<\/em>. The intensity on the other side of the lens is maximum at a distance (c) 2 f<\/em><\/p>\n\n\n\n Since the object is placed at 2 f<\/em>, the image of the object will be formed at distance of 2 f<\/em> from a converging lens. 1v-1u=1fHere, u<\/em> = \u2212 2 f<\/em> and f<\/em> = f<\/em> 1v-1-2f=1f\u21d21v=1f-12f=12fTherefore, image distance v <\/em>= 2 f<\/em><\/p>\n\n\n\n A parallel beam of light is incident on a converging lens parallel to its principal axis. As one moves away from the lens on the other side on its principal axis, the intensity of light (d) first increases then decreases<\/p>\n\n\n\n Since all beams of light are parallel to the principal axis, they will converge at the focus of a converging lens. Thus, the intensity of light increases till one reaches the focus and then starts decreasing as one moves beyond it.<\/p>\n\n\n\n A symmetric double convex lens in cut in two equal parts by a plane perpendicular to the principal axis. If the power of the original lens was 4 D, the power a cut-lens will be (a) 2 D<\/p>\n\n\n\n The lens is cut into two equal parts by a plane perpendicular to the principal axis. Thus, the radius of curvature of both the lenses become half of its original value. A symmetric double convex lens is cut in two equal parts by a plane containing the principal axis. If the power of the original lens was 4 D, the power of a divided lens will be (c) 4 D<\/p>\n\n\n\n As the lens is cut along the principal axis, the two new lens will act as two double convex lenses. Even after cutting the lens, the radius of the curvature of the two new lens will remain the same as that of the original lens, with the power equal to the original lens.<\/p>\n\n\n\n Two concave lenses L<\/em>1<\/sub> and L<\/em>2<\/sub> are kept in contact with each other. If the space between the two lenses is filled with a material of smaller refractive index, the magnitude of the focal length of the combination (c) increases<\/p>\n\n\n\n The focal length of the combination will increase. For finding out the combination of lens we have the formula:<\/p>\n\n\n\n 1F=1f1+1f2-df1f2Where, F <\/em>is the focal length for the combination 1F=1f1+1f2 F = f1f2f1+f2Hence, the focal length will increase.<\/p>\n\n\n\n A thin lens is made with a material having refractive index<\/p>\n\n\n\n \u03bc=1\u00b75. Both the side are convex. It is dipped in water (<\/p>\n\n\n\n \u03bc=1\u00b733). It will behave like (a) a convergent lens<\/p>\n\n\n\n Since the refractive index of lens is more than that of of water and both the sides are convex, it will act like a normal convergent lens (though its focal length will change).<\/p>\n\n\n\n A convex lens is made of a material having refractive index<\/p>\n\n\n\n 1\u00b72. Both the surfaces of the lens are convex. If it is dipped into water (<\/p>\n\n\n\n \u03bc=1\u00b733), it will behave like (b) a divergent lens<\/p>\n\n\n\n Since the refractive index of lens is less than that of of water and both the sides are convex, it will act like a normal diverging lens (though its focal length will change to negative).<\/p>\n\n\n\n A point object O<\/em> is placed on the principal axis of a convex lens of focal length f<\/em> = 20 cm at a distance of 40 cm to the left of it. The diameter of the lens is 10 cm. An eye is placed 60 cm to right of the lens and a distance h<\/em> below the principal axis. The maximum value of h<\/em> to see the image is (b) 2.5 cm<\/p>\n\n\n\n As the focal length of the lens is 20 cm and object distance is 40 cm from the lens, the image is formed at the centre of curvature at the right side of the lens. tan\u03b1=ABBCtan\u03b1=5 40\u03b1=tan-15 40and from right angled triangle, we have:<\/p>\n\n\n\n tan\u03b1=DEDCtan\u03b1=h20as DB-BC=DC=20 cmPutting the value of angle alpha, we get:<\/p>\n\n\n\n tantan-15 40=h205 40=h20\u21d2h=2.5 cm<\/p>\n\n\n\n The rays of different colours fail to converge at a point after going through a converging lens. This defect is called (d) chromatic aberration<\/p>\n\n\n\n When light rays of different colours do not converge at the same point after passing through a converging lens, it is called chromatic aberration. This happens because a lens has different refractive indices for different colours, i.e, for different wavelengths of light.<\/p>\n\n\n\n A concave mirror having a radius of curvature 40 cm is placed in front of an illuminated point source at a distance of 30 cm from it. Find the location of the image.<\/p>\n\n\n\n Using sign conventions, given, Using the mirror equation,<\/p>\n\n\n\n 1v+1u=2R\u21d2 1v=2R-1u\u21d21v=2-40-1-30=1-20+130\u21d21v=-30+2030\u00d720=-1030\u00d720\u21d21v=-160or, v<\/em> = \u2212 60 cm A concave mirror forms an image of 20 cm high object on a screen placed 5.0 m away from the mirror. The height of the image is 50 cm. Find the focal length of the mirror and the distance between the mirror and the object.<\/p>\n\n\n\n Given, Since, we know that<\/p>\n\n\n\n -vu=h2h1or -(-500)u=5020Where \u2018u<\/em>\u2018 is the distance of object from screen. or u=-500\u00d725=-200 cm=-2.0 mUsing mirror formula,<\/p>\n\n\n\n 1v+1u=1for 1-5+1-2=1f<\/p>\n\n\n\n or -1f=710or f=-107=-1.44 mHence, the required focal length of the concave mirror is 1.44 m.<\/p>\n\n\n\n A concave mirror has a focal length of 20 cm. Find the position or positions of an object for which the image-size is double of the object-size.<\/p>\n\n\n\n Using sign conventions, given, m=-vu=2\u21d2 v<\/em> = \u22122u<\/em><\/p>\n\n\n\n Case I (Virtual image):<\/p>\n\n\n\n Using mirror formula,<\/p>\n\n\n\n 1v+1u=1f\u21d2 12u-1u=1f\u21d2 12u=1f\u21d2 u=f2=10 cmCase II (Real image)<\/p>\n\n\n\n Using mirror formula,<\/p>\n\n\n\n 1v+1u=1f\u21d2 -12u-1u=1f\u21d2 32u=1f\u21d2 u=3f2=30 cmHence, the required positions of objects are 10 cm or 30 cm from the concave mirror.<\/p>\n\n\n\n A 1 cm object is placed perpendicular to the principal axis of a convex mirror of focal length 7.5 cm. Find its distance from the mirror if the image formed is 0.6 cm in size.<\/p>\n\n\n\n Given, 152 cmMagnification is given as,<\/p>\n\n\n\n m=-vu=hih0=0.6 Using mirror equation,1v+1u=1f\u21d2 10.6u-1u=1f\u21d2 53u-1u=215\u21d2 u=5 cmHence, the distance of the object from the mirror is 5 cm.<\/p>\n\n\n\n A candle flame 1.6 cm high is imaged in a ball bearing of diameter 0.4 cm. If the ball bearing is 20 cm away from the flame, find the location and the height of the image.<\/p>\n\n\n\n Given, Using mirror formula,<\/p>\n\n\n\n 1v+1u=2RPutting the values according to sign conventions, we get,<\/p>\n\n\n\n 1(-20)+1v=20.2\u21d2 1v=120+10\u21d2 v=0.1 cm or 1.0 mm inside the ball bearing.Magnification=m=A\u2019B\u2019AB=-vu=m=A\u2019B\u2019200=1200\u21d2 A\u2019B\u2019=AB200=+1.6200=+0.08 cm (+0.008 cm)Hence, the distance of the image is 1 cm. A 3 cm tall object is placed at a distance of 7.5 cm from a convex mirror of focal length 6 cm. Find the location, size and nature of the image.<\/p>\n\n\n\n Given, 1v+1u=1f\u21d2 1v=1f-1uPutting values according to sign convention,<\/p>\n\n\n\n 1v=16-1(-7.5)1v=16+17.5=310 \u21d2v=103 cmMagnification = m<\/em> =<\/p>\n\n\n\n -vu=107.5\u00d73<\/p>\n\n\n\n \u21d2 A\u2019B\u2019AB=107.5\u00d73\u21d2 A\u2019B\u2019=10075=43=1.33 cmWhere A\u2019B\u2019 is the height of the image.<\/p>\n\n\n\n Hence, the required location of the image is<\/p>\n\n\n\n 103 cmfrom the pole and image height is 1.33 cm. Nature of the image is virtual and erect.<\/p>\n\n\n\n A U-shaped wire is placed before a concave mirror having radius of curvature 20 cm as shown in figure. Find the total length of the image. Given, R2= \u221210 cm Therefore, u<\/em> = \u221240 cm \u21d2 1v+1u=1f<\/p>\n\n\n\n 1v=-110-1-40=-340<\/p>\n\n\n\n \u21d2 v=-403=13.3 cm<\/p>\n\n\n\n So, PB\u2019=13.3 cm<\/p>\n\n\n\n m=A\u2019B\u2019AB=-(v)u<\/p>\n\n\n\n =-(-13.3)-40=-13\u21d2 A\u2019B\u2019=-103=-3.33 cmFor part CD of the U shaped wire, PC = 30 cm 1v=1f-1u<\/p>\n\n\n\n 1v=-110-1-301v=-110+130=-115\u21d2 v<\/em> = \u221215 cm = PC\u2019<\/p>\n\n\n\n m=C\u2019D\u2019CD=-vu<\/p>\n\n\n\n =-(-15)-30=-12\u21d2 C\u2019D\u2019 = 5 cm Hence, the total length of the U- shaped wire is A\u2019B\u2019 + B\u2019C\u2019 + C\u2019D\u2019 A man uses a concave mirror for shaving. He keeps his face at a distance of 25 cm from the mirror and gets an image which is 1.4 times enlarged. Find the focal length of the mirror.<\/p>\n\n\n\n Given, 1f=1v+1u<\/p>\n\n\n\n \u21d2 1f=135-125<\/p>\n\n\n\n =5-7175=-2175\u21d2 f<\/em> = \u221287.5<\/p>\n\n\n\n Hence, the required focal length of the concave mirror is 87.5 cm.<\/p>\n\n\n\n If the light moving in a straight line bends by a small but fixed angle, it may be a case of (a) reflection When the light strikes on a surface nearly parallel to it, it then bends by a small and fixed angle after reflection. Also, when the light travels from one medium to another with slight differences in their refractive indices, it bends by a small angle. Thus, the bending of light by a small but fixed angle can be the case of either reflection or refraction.<\/p>\n\n\n\n Mark the correct options. (b) If the final rays are converging, we have a real image.<\/p>\n\n\n\n This is because a real image is formed by converging reflected\/refracted rays from a mirror\/lens.<\/p>\n\n\n\n Which of the following (referred to a spherical mirror) do (does) not depend on whether the rays are paraxial or not? (a) Pole Paraxial rays are the light rays close to the principal axis. The focus of the spherical mirror for paraxial rays are different from the focus for marginal rays. So, the focus depends on whether the rays are paraxial or marginal. \u00e2\u20ac\u2039The pole, radius of curvature and the principal axis of a spherical mirror do not depend on paraxial or marginal rays.<\/p>\n\n\n\n The image of an extended object, placed perpendicular to the principal axis of a mirror, will be erect if (c) the object is real but the image is virtual The virtual image of a real object and the real image of a virtual object are always erect.<\/p>\n\n\n\n A convex lens forms a real image of a point object placed on its principals axis. If the upper half of the lens is painted black, (c) the image will not be shifted If the upper half portion of the convex lens is painted, then only the intensity of the image will decrease, as the amount of light passing through the lens will decrease. Also, there will be no shift in the position of the image because all the parameters remain the same.<\/p>\n\n\n\n Consider three converging lenses L<\/em>1<\/sub>, L<\/em>2<\/sub> and L<\/em>3<\/sub> having identical geometrical construction. The index of refraction of L<\/em>1<\/sub> and L<\/em>2<\/sub> are<\/p>\n\n\n\n \u03bc1 and \u03bc2respectively. The upper half of the lens L<\/em>3<\/sub> has a refractive index<\/p>\n\n\n\n \u03bc1and the lower half has<\/p>\n\n\n\n \u03bc2(figure 18-Q3). A point object O<\/em> is imaged at O<\/em>1<\/sub> by the lens L<\/em>1<\/sub> and at O<\/em>2<\/sub> by the lens L<\/em>2<\/sub> placed in same position. If L<\/em>3<\/sub> is placed at the same place, (a) there will be an image at O<\/em>1<\/sub> The lens L<\/em>1<\/sub> converges the light at point O<\/em>1<\/sub> and lens L<\/em>2<\/sub> converges the light at O<\/em>2<\/sub>. As the upper half of lens L<\/em>
For a large refractive index, the value of the critical angle is small. Thus, for the diamond, the critical angle for total internal reflection is much smaller than that of the glass. Therefore, a great percentage of incident light gets internally reflected several times before it emerges out.<\/p>\n\n\n\nQuestion 4:<\/h4>\n\n\n\n
Figure<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 5:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 6:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Instead, if the paper is placed at the position of a real image, it will burn as the light rays are actually concentrating on the point. Also, when we have the real image of a virtual source, it will also cause the burning of paper due to same reason; the light rays are actually meeting at the point.<\/p>\n\n\n\nQuestion 7:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 8:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 9:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 10:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/12\/5781.png\")
Let actual height be h <\/em>and the apparent height be h<\/em>\u2018.
Here, the refraction is taking place from rarer to denser medium and a virtual image is formed.
Using<\/p>\n\n\n\n
u<\/em> and v<\/em> are object and image distances, respectively.
R <\/em>is the radius of curvature, here we will take it as \u221e.<\/p>\n\n\n\n
v<\/em> = u<\/em> \u00d7 <\/em>\u03bc<\/em>2<\/sub>
We know magnification is given by:<\/p>\n\n\n\nQuestion 11:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 12:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
\u03bc<\/em>2<\/sub> and \u03bc<\/em>1<\/sub> are refractive indexes of lens and air respectively.
R<\/em>1<\/sub> and R<\/em>2<\/sub> are the radius of curvature of convex and plane surfaces respectively.
Here, R<\/em>2 <\/sub>= \u221e because of the plane surface.<\/p>\n\n\n\n
When the convex surface is facing the object, we have:<\/p>\n\n\n\nQuestion 13:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 14:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 15:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
To reduce the aperture of a parallel beam of light without losing the energy of the light and also increase the intensity, we have to place the lens L2<\/sub> within the focal range of L1.<\/sub>![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/12\/Q15_SA.png\")
<\/p>\n\n\n\nQuestion 16:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 17:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 18:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 19:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 1:<\/h4>\n\n\n\n
(a) All the reflected rays meet at a point when produced backward.
(b) Only the reflected rays close to the normal meet at a point when produced backward.
(c) Only the reflected rays making a small angle with the mirror meet at a point when produced backward.
(d) Light of different colours make different images.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Page No 411:<\/h4>\n\n\n\n
Question 2:<\/h4>\n\n\n\n
(a) light goes from optically rarer medium (smaller refractive index) to optically denser medium
(b) light goes from optically denser medium to rarer medium
(c) the refractive indices of the two media are close to each other
(d) the refractive indices of the two media are widely different.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 3:<\/h4>\n\n\n\n
(a) are easy to handle geometrically
(b) contain most of the intensity of the incident light
(c) from nearly a point image of a point source
(d) show minimum dispersion effect.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 4:<\/h4>\n\n\n\n
(a) infinity
(b) pole
(c) focus
(d) 15 cm behind the mirror.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
v<\/em> = ?
f<\/em> = 30 cm<\/p>\n\n\n\nQuestion 5:<\/h4>\n\n\n\n
(a) is plane
(b) is convex
(c) is concave
(d) may be any spherical mirror.
Figure<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 6:<\/h4>\n\n\n\n
(a) is always real
(b) is always virtual
(c) is certainly real if the object is virtual
(d) is certainly virtual if the object is real.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 7:<\/h4>\n\n\n\n
(a) the image forms between O<\/em>\u2018 and O<\/em>\u201d
(b) the image forms to the left of O<\/em>\u2018
(c) the image forms to the right of O<\/em>\u201d
(d) two images form, one at O<\/em>\u2018 and the other at O<\/em>\u201c.
Figure<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 8:<\/h4>\n\n\n\n
(a)<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Thus, u<\/em>new<\/sub> = u<\/em> + t<\/em> and v<\/em>new<\/sub> = v<\/em> \u2212 t<\/em>
We get the new lens formula:<\/p>\n\n\n\nQuestion 9:<\/h4>\n\n\n\n
(a)<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 10:<\/h4>\n\n\n\n
(a) f<\/em>
(b) between f<\/em> and 2 f<\/em>
(c) 2 f<\/em>
(d) more than 2 f<\/em>.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
It can also be shown from the lens formula:<\/p>\n\n\n\n
On putting the respective values we get:<\/p>\n\n\n\nQuestion 11:<\/h4>\n\n\n\n
(a) remains constant
(b) continuously increases
(c) continuously decreases
(d) first increases then decreases.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 12:<\/h4>\n\n\n\n
(a) 2 D
(b) 3 D
(c) 4 D
(d) 5 D.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
As,
f=R2P=1fRadius of curvature becomes half. Therefore, the focal also reduces to half its original value.
Thus, the power (P<\/em>) of the two cut-lenses will be equal.
2 P<\/em> = 4 D
P<\/em> = 2 D<\/p>\n\n\n\nQuestion 13:<\/h4>\n\n\n\n
(a) 2 D
(b) 3 D
(c) 4 D
(d) 5 D.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 14:<\/h4>\n\n\n\n
(a) becomes undefined
(b) remains unchanged
(c) increases
(d) decreases.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
d<\/em> is the separation between two lenses
Here, d<\/em> = 0<\/p>\n\n\n\nQuestion 15:<\/h4>\n\n\n\n
(a) a convergent lens
(b) a divergent lens
(c) a rectangular slab
(d) a prism.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 16:<\/h4>\n\n\n\n
(a) a convergent lens
(b) a divergent lens
(c) a rectangular slab
(d) a prism.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 17:<\/h4>\n\n\n\n
(a) 0
(b) 2.5 cm
(c) 5 cm
(d) 10 cm.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/12\/Objective-1_Q17.png\")
From right angled triangle ABC,<\/p>\n\n\n\nQuestion 18:<\/h4>\n\n\n\n
(a) spherical aberration
(b) distortion
(c) coma
(d) chromatic aberration.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Page No 412:<\/h4>\n\n\n\n
Question 1:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Distance of object from mirror, u<\/em> = \u2212 30 cm,
Radius of curvature of concave mirror R<\/em> = \u2212 40 cm<\/p>\n\n\n\n![\"\"\/](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/12\/1_Ans21.png\")
<\/figure>\n\n\n\n
Hence, the required image will be located at a distance of 60 cm in front of the concave mirror.<\/p>\n\n\n\nQuestion 2:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Height of the object, h<\/em>1<\/sub> = 20 cm,
Distance of image from screen v<\/em> = \u22125.0 m = \u2212500 cm,<\/p>\n\n\n\n![\"\"\/](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/12\/2_Ans12.png\")
<\/figure>\n\n\n\n
(As the image is inverted)<\/p>\n\n\n\nQuestion 3:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Focal length of the concave mirror:
f<\/em> = \u221220 cm
As per the question,
Magnification (m<\/em>) is:<\/p>\n\n\n\n![\"\"\/](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/12\/3_Ans14.png\")
<\/figure>\n\n\n\nQuestion 4:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Height of the object, h<\/em>1<\/sub> = 1 cm
Focal length of the concave mirror, f<\/em> = 7.5 cm =<\/p>\n\n\n\nQuestion 5:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Height (h<\/em>1<\/sub>) of the candle flame taken as object AB<\/em> = 1.6 cm
Diameter of the ball bearing (d<\/em>) = 0.4 cm
So, radius = 0.2 cm
Distance of object, u<\/em> = 20 cm<\/p>\n\n\n\n![\"\"\/](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/12\/5_Ans16.png\")
<\/figure>\n\n\n\n
Height of the image is 0.008 cm.<\/p>\n\n\n\nQuestion 6:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Height of the object AB<\/em>, h<\/em>1<\/sub> = 3 cm
Distance of the object from the convex mirror, u<\/em> = \u22127.5 cm
Focal length of the convex mirror (f<\/em>) = 6 cm
Using mirror formula,<\/p>\n\n\n\n![\"\"\/](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/12\/6_Ans9.png\")
<\/figure>\n\n\n\nQuestion 7:<\/h4>\n\n\n\n
Figure<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Radius of curvature of concave mirror, R<\/em> = 20 cm
So its focal length will be f<\/em> =<\/p>\n\n\n\n
For part AB of the U shaped wire, PB = 30 + 10 = 40 cm<\/p>\n\n\n\n![\"\"\/](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/12\/7_Ans13.png\")
<\/figure>\n\n\n\n
Using mirror formula,<\/p>\n\n\n\n
Therefore, u<\/em> = \u221230 cm
Again, using mirror equation:<\/p>\n\n\n\n
\u21d2 B\u2019C\u2019 = PC\u2019 \u2212 PB\u2019
= 15 \u2212 13.3 = 1.7 cm<\/p>\n\n\n\n
= (3.3) + (1.7) + 5 = 10 cm<\/p>\n\n\n\nQuestion 8:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Distance of the man\u2019s face (here, taken as object), u<\/em> = \u221225 cm
According to the question, magnification, m<\/em> = 1.4
m=A\u2019B\u2019AB=-vu\u21d2 1.4=-(v)-25\u21d2 1410=v25\u21d2 v=25\u00d71410=35 cmUsing equation of mirror, we get:<\/p>\n\n\n\nQuestion 1:<\/h4>\n\n\n\n
(a) reflection
(b) refraction
(c) diffraction
(d) dispersion.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
(b) refraction<\/p>\n\n\n\nQuestion 2:<\/h4>\n\n\n\n
(a) If the incident rays are converging, we have a real object.
(b) If the final rays are converging, we have a real image.
(c) The image of a virtual object is called a virtual image.
(d) If the image is virtual, the corresponding object is called a virtual object.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 3:<\/h4>\n\n\n\n
(a) Pole
(b) Focus
(c) Radius of curvature
(d) Principal axis<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
(c) Radius of curvature
(d) Principal axis<\/p>\n\n\n\nQuestion 4:<\/h4>\n\n\n\n
(a) the object and the image are both real
(b) the object and the image are both virtual
(c) the object is real but the image is virtual
(d) the object is virtual but the image is real.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
(d) the object is virtual but the image is real<\/p>\n\n\n\nQuestion 5:<\/h4>\n\n\n\n
(a) the image will be shifted downward
(b) the image will be shifted upward
(c) the image will not be shifted
(d) the intensity of the image will decrease.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
(d) the intensity of the image will decrease<\/p>\n\n\n\nQuestion 6:<\/h4>\n\n\n\n
(a) there will be an image at O<\/em>1<\/sub>
(b) there will be an image at O<\/em>2<\/sub>.
(c) the only image will form somewhere between O<\/em>1<\/sub> and O<\/em>2<\/sub>
(d) the only image will form away from O<\/em>2<\/sub>.
Figure<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
\u00e2\u20ac\u2039(b) there will be an image at O<\/em>2<\/sub><\/p>\n\n\n\n