Ray Optics MCQ Class XII.(TN State Board) pptx

Physics Multiple choice questions and answers with explanation. (Class XII Physics TN State board)

1. Ray Optics MCQ Class-XII (TN State Board) (With explanation)

2. By Dr M. Arunachalam Head, Department of Physics ( Rtd .) Sri SRNM College, Sattur

3. 1. The speed of light in an isotropic medium depends on, (a) its intensity (b)its wavelength (c) the nature of propagation (d) the motion of the source w.r.t medium We know that n= c air / c medium ie . c medium = c air /n n - Refractive index. c air - Speed of light in air c medium - Speed of light in the medium

4. Solution From the concept of dispersion, n depends on wavelength λ . Hence, t he speed of light in an isotropic medium depends on , its wavelength Ans : b

5. 2. A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a way that its end closer to the pole is 20 cm away from the mirror. The length of the image is, 2.5 cm (b) 5cm (c) 10 cm (d) 15cm For a concave mirror, 1/u + 1/v = 1/f u – Distance between the mirror and the object v – Distance between the mirror and the image f – Focal length of the mirror

6. Contd … For end A u 1 = -20 cm. So we can write 1/-20 + 1/v 1 = 1/-10 ie 1/v 1 = 1/-10 -1/-20 = 1/-10 + 1/20 ie . 1/v 1 = 1/20-1/10 = -1/20 ie . v 1 = -20cm . For end B u 2 = -30 cm. (Since the length of the rod is 10cm.) 1/-30 + 1/v 2 = 1/-10 ie 1/v 2 = 1/-10 -1/-30 = 1/30 +1/-10 ie . 1/v 2 = (1-3)/30 = -2/30 = -1/15 ie . v 2 = -15cm . Length of the image = -15 – (-20) = -15 + 20 = 5cm. Ans: b

7. 3. An object is placed in front of a convex mirror of focal length f . The maximum and minimum distance of an object from the mirror such that the image formed is real and magnified. 2 f and c (b) c and ∞ (c) f and O (d) None of these By its character, convex mirror will not form real images. Hence the answer is d ie . None of these

8. 4. For light incident from air on a slab of refractive index 2, the maximum possible angle of refraction is, (a) 30 o (b) 45 o (c) 60 o (d) 90 o By Snell’s law, n = sin i / sin r i – Angle of incidence r – Angle of refraction n - Refractive index = 2

9. Contd … By Snell’s law, when i increases r will also increase. We will get the maximum possible r value ( r max ) for maximum value of i ( ie 90 o ) Now we have n = sin i / sin r ie . 2 = sin 90 o / sin r ie . . 2 = 1/ sin r ie . sin r = 1/2 ie r = sin -1 (1/2) = 30 o The maximum possible angle of refraction = 30 o Ans: a

10. 5. If the velocity and wavelength of light in air is Va and λa and that in water is Vw and λw , then the refractive index of water is, (a)V w / V a (b) V a/ V w (c) λ a / λ w (d) Va λ a / Vw λ w Refractive index is n = velocity of light in air ( Va)/ velocity of light in medium ( Vmed .) ie . n = V a/ V w Ans: b

11. 6. Stars twinkle due to, (a) reflection (b) total internal reflection (c) refraction (d) polarisation There are different layers in the atmosphere with varying density and temperature that change with time. This causes variation of refractive index in the atmosphere with time. So the stars twinkle due to atmospheric refraction. Ans: c

12. 7. When a biconvex lens of glass having refractive index 1.47 is dipped in a liquid, it acts as a plane sheet of glass. This implies that the liquid must have refractive index, (a) less than one (b) less than that of glass (c) greater than that of glass (d) equal to that of glass For the convex lens to act as the plane sheet, the liquid must have refractive index equal to that of glass . Ans: d

13. 8. The radius of curvature of curved surface at a thin planoconvex lens is 10 cm and the refractive index is 1.5. If the plane surface is silvered, then the focal length will be, 5 cm (b) 10 cm (c) 15 cm (d) 20 cm We know that, 1/f = [(n 2 - 1)/n] [ 1/R 1 - 1/R 2 ] R 2 - Radius curvature of the plane surface = So the silvering of the plane surface will not make any impact on f Hence focal length f = 10 cm. Ans: b  

14. 9. An air bubble in glass slab of refractive index 1.5 (near normal incidence) is 5 cm deep when viewed from one surface and 3 cm deep when viewed from the opposite face. The thickness of the slab is, 8 cm (b) 10 cm (c) 12 cm (d) 16 cm For I surface n = d 1 /d 1 A n – Refractive index = 1.5 d 1 – Real depth d 1 A - Apparent depth = 5cm. ie . d 1 - nd 1 A = 1.5x5 = 7.5cm.

15. Contd … For II surface n = d 2 /d 2 A n – Refractive index = 1.5 d 2 – Real depth d 2 A - Apparent depth = 3cm. ie . d 2 - nd 2 A = 1.5x3 = 4.5cm. Thickness of the slab is d 2 + d 2 = 7.5 + 4.5 =12cm Thickness of the slab = 12cm Ans: c

16. 10. A ray of light travelling in a transparent medium of refractive index n falls, on a surface separating the medium from air at an angle of incidents of 45 o . The ray can undergo total internal reflection for the following n , n = 1.25 (b) n = 1.33 (c) n = 1.4 (d) n = 1.5 The condition for total internel reflection is, the angle of incidene is i > C (critical angle). Sin C = 1/n sin i > sin C ie. sin i > 1/n. Here the angle of incidence i = 45 ie . n > 1/  

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