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Ray Optics and Optical Instruments

Multiple Choice Questions

391.
An object is located 4 m from the first of two thin converging lenses of focal lengths 2 m and 1 m, respectively. The lenses are separated by 3 m. The final image formed by the second lens is located from the source at a distance of

8.0 m

5.5 m

6.0 m

6.5 m

5.5 m

From the lens formula (for first lens)

$\frac{1}{f_{1}} = \frac{1}{v_{1}} - \frac{1}{u_{1}} \Rightarrow \frac{1}{2} = \frac{1}{v_{1}} - \frac{1}{(- 4)} \Rightarrow \frac{1}{2} + \frac{1}{4} = \frac{1}{v_{1}} = \frac{3}{4} . . . . . . . . (i) \Rightarrow v_{1} = \frac{4}{3}, u_{2} = 3 - \frac{4}{3} \Rightarrow v_{1} = 5 / 3$

This image will be treated as the source for second lens, then again from lens formula, we have

$\frac{1}{f_{2}} = \frac{1}{v_{2}} - \frac{1}{u_{2}} \Rightarrow \frac{1}{1} = \frac{1}{v_{2}} + 5 / 3 [By Eq . (i)] \Rightarrow 1 - 5 / 3 = \frac{1}{v_{2}} \Rightarrow v_{2} = - 3 / 2$

This is the final image distance from 2nd lens. So, the overall distance of image from the primary source (or object)

Let d = 4 + 3 − 1.5 = 5.5

392.

A ray of light is incident at an angle i on a glass slab of refractive index µ. The angle between reflected and refracted light is 90°. Then, the relationship between i and µ is