Question 155: Ray Optics

Question

A concave mirror has radius of curvature of 40 cm. It is at the bottom of a glass that has water filled up to 5
cm (see figure). If a small particle is floating on the surface of water, its image as seen, from directly above
the glass, is at a distance d from the surface of water. The value of d is dose to: (Refractive index of water =
1.33)
![](https://imagex.cdn.examgoal.net/3rsa0w2w9jx3on8ko/f002fe1e-5084-4d01-8cc5-76dc2cbc53dd/e9764a80-92cf-11e9-86c1-73f7f991b766/file-3rsa0w2w9jx3on8kp.png)

Accepted Answer

Light incident from particle P will be reflected at mirror.
u = 5cm, f = $m - {R \over 2} =  - 20cm$
${1 \over v} + {1 \over u} = {1 \over f};$   ${v_1} =  + {{20} \over 3}cm$
This image will act as object for light getting refracted at
water surface.
So, object distance $d = 5 + {{20} \over 3} = {{35} \over 3}cm$
Below water surface.
After refraction, final image is at
$d' = d\left( {{{{\mu _2}} \over {{\mu _1}}}} \right) = \left( {{{35} \over 3}} \right)\left( {{1 \over {4/3}}} \right)$
= ${{35} \over 4} = 8.75\,cm$
$ \approx $ 8.8 cm

Answer

11.7 cm

Answer

6.7 cm

Answer

13.4 cm

Answer

8.8 cm

Ray Optics

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