JEE MAIN - Physics (2019 - 12th April Morning Slot - No. 3)
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)
_12th_April_Morning_Slot_en_3_1.png)
11.7 cm
6.7 cm
13.4 cm
8.8 cm
Explanation
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
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
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