CBSE 12th Standard Physics Subject Ray Optics and Optical Instruments Chapter Case Study Questions 2021
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CBSE 12th Standard Physics Subject Ray Optics and Optical Instruments Case Study Questions 2021
12th Standard CBSE
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Reg.No. :
Physics
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A convex or converging lens is thicker at the centre than at the edges. It converges a parallel beam of light on refraction through it. It has a real focus. Convex lens is of three types:
(i) Double convex lens
(ii) Plano-convex lens
(iii) Concavo-convex lens. Concave lens is thinner at the centre than at the edges. It diverges a parallel beam of light on refraction through it. It has a virtual focus.
(i) A point object 0 is placed at a distance of 0.3 m from a convex lens (focal length 0.2 m) cut into two halves each of which is displaced by 0.0005 m as shown in figure.What will be the location of the image?
(a) 30 cm right of lens (b) 60 ern right of lens (c) 70 ern left of lens (d) 40 cm left oflens (ii) Two thin lenses are in contact and the focal length of the combination is 80 cm. If the focal length of one lens is 20 cm, the focal length of the other would be
(a) -26.7 cm (b) 60 crn (c) 80 cm (d) 20 cm (iii) A spherical air bubble is embedded in a piece of glass. For a ray of light passing through the bubble, it behaves like a
(a) converging lens (b) diverging lens (c) plano-converging lens (d) plano-diverging lens (iv) Lens used in magnifying glass is
(a) Concave lens (b) Convex lens (c) Both (a) and (b) (d) None of the above (v) The magnification of an image by a convex lens is positive only when the object is placed
(a) at its focus F (b) between F and 2F (c) at 2F (d) between F and optical centre -
Refraction of light is the change in the path oflight as it passes obliquely from one transparent medium to another medium. According to law of refraction \(\frac{\sin i}{\sin r}={ }^{1} \mu_{2}\) where \({ }^{1} \mu_{2}\) is called refractive index of second medium with respect to first medium. From refraction at a convex spherical surface, we have \(\frac{\mu_{2}}{v}-\frac{\mu_{1}}{u}=\frac{\mu_{2}-\mu_{1}}{R}\) Similarly from refraction at a concave spherical surface when object lies in the rarer medium, we have \(\frac{\mu_{2}}{v}-\frac{\mu_{1}}{u}=\frac{\mu_{2}-\mu_{1}}{R}\) and when object lies in the denser medium, we have \(\frac{\mu_{1}}{v}-\frac{\mu_{2}}{u}=\frac{\mu_{1}-\mu_{2}}{R}\).
(i) Refractive index of a medium depends upon(a) nature of the medium (b) wavelength of the light used (c) temperature (d) all of these (ii) A ray of light of frequency 5 x 1014 Hz is passed through a liquid. The wavelength of light measured inside the liquid is found to be 450 x 10-9 m. The refractive index of the liquid is
(a) 1.33 (b) 2.52 (c) 2.22 (d) 0.75 (iii) A ray of light is incident at an angle of 60° on one face of a rectangular glass slab of refractive index 1.5. The angle of refraction is
(a) sin-1(0.95) (b) sin-1(0.58) (c) sin-1(0.79) (d) sin-1(0.86) (iv) A point object is placed at the centre of a glass sphere of radius 6 cm and refractive index 1.5. The distance of the virtual image from the surface of sphere is
(a) 2 cm (b) 4 cm (c) 6 cm (d) 12 cm (v) In refraction, light waves are bent on passing from one medium to the second medium because in the second medium
(a) the frequency is different (b) the co-efficient of elasticity is different (c) the speed is different (d) the amplitude is smaller. -
The lens maker's formula relates the focal length of a lens to the refractive index of the lens material and the radii of curvature of its two surfaces. This formula is called so because it is used by manufacturers to design lenses of
required focal length from a glass of given refractive index. If the object is placed at infinity, the image will be formed at focus for both double convex lens and double concave lens
Therefore, lens maker's formula is \(\frac{1}{f}=\left[\frac{\mu_{2}-\mu_{1}}{\mu_{1}}\right]\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]\)
When lens is placed in air, \(\mu\) 1 = 1 and \(\mu\)2 = \(\mu\). The lens maker formula takes the form \(\frac{1}{f}=(\mu-1)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]\)
(i) The radius of curvature of each face of biconcave lens with refractive index 1.5 is 30 cm. The focal length of the lens in air is(a) 12 cm (b) 10 cm (c) 20 cm (d) 30 cm (ii) The radii of curvature of the faces of a double convex lens are 10 cm and 1.5 cm. If focal length is 12 cm, then refractive index of glass is
(a) 1.5 (b) 1.78 (c) 2.0 (d) 2.52 (iii) An under-water swimmer cannot see very clearly even in absolutely clear water because of
(a) absorption oflight in water (b) scattering of light in water (c) reduction of speed of light in water (d) change in the focal length of eye-lens (iv) A thin lens of glass (\(\mu\) = 1.5) offocallength 10 cm is immersed in water (\(\mu\) = 1.33). The new focal length is
(a) 20 cm (b) 40 cm (c) 48 cm (d) 12 cm (v) An object is immersed in a fluid. In order that the object becomes invisible, it should
(a) behave as a perfect reflector (b) absorb all light falling on it (c) have refractive index one (d) have refractive index exactly matching with that of the surrounding fluid. -
A prism is a portion of a transparent medium bounded by two plane faces inclined to each other at a suitable angle. A ray of light suffers two refractions on passing through a prism and hence deviates through a certain angle from its original path. The angle of deviation of a prism is, \(\delta\) = (\(\mu\)- 1) A, through which a ray deviates on passing through a thin prism of small refracting angle A.
If \(\mu\) is refractive index of the material of the prism, then prism formula is,\(\mu=\frac{\sin \left(A+\delta_{m}\right) / 2}{\sin A / 2}\)
(i) For which colour, angle of deviation is minimum?(a) Red (b) Yellow (c) Violet (d) Blue (ii) When white light moves through vacuum
(a) all colours have same speed (b) different colours have different speeds (c) violet has more speed than red (d) red has more speed than violet. (iii) The deviation through a prism is maximum when angle of incidence is
(a) 45° (b) 70° (c) 90° (d) 60° (iv) What is the deviation produced by a prism of angle 6°? (Refractive index of the material of the prism is 1.644).
(a) 3.864° (b): 4.595° (c) 7.259° (d) 1.252° (v) A ray of light falling at an angle of 50° is refracted through a prism and suffers minimum deviation. If the angle of prism is 60°, then the angle of minimum deviation is
(a) 45° (b) 75° (c) 50° (d) 40° -
An optical fibre is a thin tube of transparent material that allows light to pass through, without being refracted into the air or another external medium. It make use of total internal reflection. These fibres are fabricated in such a way that light reflected at one side of the inner surface strikes the other at an angle larger than critical angle. Even, if fibre is bent, light can easily travel along the length.
(i) Which of the following is based on the phenomenon of total internal reflection of light?(a) Sparkling of diamond (c) Instrument used by doctors for endoscopy (b) Optical fibre communication (d) All of these (ii) A ray of light will undergo rotal internal reflection inside the optical fibre, if it
(a) goes from rarer medium to denser medium (b) is incident at an angle less than the critical angle (c) strikes the interface normally (d) is incident at an angle greater than the critical angle (iii) If in core, angle of incidence is equal to critical angle, then angle of refraction will be
(a) 0° (b) 45° (c) 90 (d) 180° (iv) In an optical fibre (shown), correct relation for refractive indices of core and cladding is
(a) n1 = n2 (b) n1 > n2 (c) n1 < n2 (d) n1 + n2 = 2 (v) If the value of critical angle is 30° for total internal reflection from given optical fibre, then speed of light in that fibre is
(a) 3 x 108 m S-1 (b) 1.5 x 108 m S-1 (c) 6 x 108 m s-1 (d) 4.5 x 108 m s-1
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CBSE 12th Standard Physics Subject Ray Optics and Optical Instruments Case Study Questions 2021 Answer Keys
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(i) (b): Each half lens will form an image in the same plane. The optic axes of the lenses are displaced
\(\frac{1}{v}-\frac{1}{(-30)}=\frac{1}{20} ; v=60 \mathrm{~cm}\)
(ii) (a): Here \(f_{1}=20 \mathrm{~cm} ; f_{2}=?\)
F= 80 cm
\(\text { As } \frac{1}{f_{1}}+\frac{1}{f_{2}}=\frac{1}{F} \Rightarrow \frac{1}{f_{2}}=\frac{1}{F}-\frac{1}{f_{1}}\)
\(\frac{1}{f_{2}}=\frac{1}{80}-\frac{1}{20}=\frac{-3}{80}\)
\(f_{2}=\frac{-80}{3}=-26.7 \mathrm{~cm}\)
(iii) (b): The bubble behaves libe a diverging lens
(iv) (b): Convex lens is used in magnifying glass.
(v) (d) -
(i) (d): Refractive index ofa medium depends upon nature and temperature of the medium, wavelength of light.
(ii) (a): Here \(v=5 \times 10^{14} \mathrm{~Hz} ; \lambda=450 \times 10^{-9} \mathrm{~m}\)
\(c=3 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}\)
Refractive index of the liquid,
\(\mu=\frac{c}{v}=\frac{c}{v \lambda}=\frac{3 \times 10^{8}}{5 \times 10^{14} \times 450 \times 10^{-9}} \)
\(\mu=1.33\)
(iii) (b): Here i = 60° ; \(\mu\)= 1.5
By snell's law, \(\mu=\frac{\sin i}{\sin r}\)
\(\sin r=\frac{\sin i}{\mu}=\frac{\sin 60^{\circ}}{1.5}=\frac{0.866}{1.5} \)
\(\sin r=0.5773 \text { or } r=\sin ^{-1}(0.58)\)
(iv) (c): As object is at the centre of the sphere, the image must be at the centre only.
\(\therefore\) Distance of virtual image from centre of sphere = 6cm.
(v) (c): Speed of light in second medium is different than that in first medium. -
(i) (d): Here, \(\mu=1.5 ; R_{1}=30 \mathrm{~cm}\)
R2 = -30 cm
\(\text { As } \frac{1}{f}=(\mu-1)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right] \)
\(=(1.5-1)\left[\frac{1}{30}-\frac{1}{-30}\right]=-0.5 \times \frac{2}{30}-\frac{-1}{30}\)
\(f=-30 \mathrm{~cm}\)
(ii) (a): Here, f= 12 cm ; R1 = 10 cm
R2 = -15 cm
\(\text { As } \frac{1}{f}=(\mu-1)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right] \)
\(\frac{1}{12}=(\mu-1)\left[\frac{1}{10}+\frac{1}{15}\right] \)
\(\mu=1.5\)
(iii) (d): The eye-lens is surrounded by a different medium than air. This will change the focal length of the eye-lens. The eye cannot accommodate all images as it would do in air.
(iv) (b): \(\frac{1}{f}=(1.5-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
\(\text { and } \frac{1}{f_{w}}=\left(\frac{1.5}{1.33}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \)
\(\frac{f_{w}}{f}=\frac{0.5 \times 1.33}{0.17}=4 \)
\(f_{w}=4 f=4 \times 10=40 \mathrm{~cm}\)
(v) (d): If the refractive index of two media are same,the surface of separation does not produce refraction or reflection which helps in visibility. -
(i) (a): Angle of deviation is minimum for the red colour.
(ii) (a): In vacuum all colours have same speed, because there is no dispersion of light in vacuum.
(iii) (c): The deviation is maximum when angle is 90°.
(iv) (a): \(A=6^{\circ} ; \mu=1.644\)
\(f=(\mu-1) A \)
\(f=(1.644-1) 6=0.644 \times 6 \)
\(\delta=3.864^{\circ}\)
(v) (d): \(i_{1}=50^{\circ} ; A=60^{\circ}, \delta_{m}=?\)
\(A+\delta_{m}=i_{1}+i_{2}=50^{\circ}+50^{\circ}=100^{\circ} \)
\(\delta_{m}=100^{\circ}-A=100-60^{\circ}=40^{\circ}\) -
(i) (d): Total internal reflection is the basis for following phenomenon:
(a) Sparkling of diamond.
(b) Optical fibre communication.
(c) Instrument used by doctors for endoscopy.
(ii) (d): Total internal reflection (TIR) is the phenomenon that involves the reflection of all the incident light off the boundary. TIR only takes place when both of the following two conditions are met:The light is in the more denser medium and approaching the less denser medium.The angle of incidence is greater than the critical angle.
(iii) (c) : If incidence of angle, i = critical angle e, then angle of refraction, r = \(90^{\circ}\)
(iv) (b): In optical fibres, core is surrounded by cladding, where the refractive index of the material of the core is higher than that of cladding to bound the light rays inside the core.
(v) (b): From Snell's law, \(\sin C={ }_{1} n_{2}=\frac{v_{1}}{v_{2}}\)
where, c = critical angle = 30° and V1 and V2 are speed oflight in medium and vacuum, respectively.
We know that, v2 = 3 x 108 m s-l
\(\therefore \quad \sin 30^{\circ}=\frac{v_{1}}{3 \times 10^{8}} \)
\(\Rightarrow \quad v_{1}=3 \times 10^{8} \times \frac{1}{2} \Rightarrow v_{1}=1.5 \times 10^{8} \mathrm{~ms}^{-1} \)
