JEE MAIN - Physics (2019 - 9th April Evening Slot - No. 25)
A thin convex lens L (refractive index = 1.5)
is placed on a plane mirror M. When a pin is
placed at A, such that OA = 18 cm, its real
inverted image is formed at A itself, as shown
in figure. When a liquid of refractive index μ1
is put between the lens and the mirror, The pin
has to be moved to A', such that OA' = 27 cm,
to get its inverted real image at A' itself. The
value of μ1 will be :-
_9th_April_Evening_Slot_en_25_1.png)
$$\sqrt 3 $$
3 / 2
$$\sqrt 2 $$
4 / 3
Explanation
For image to form at object itself, says must retrace their path back to object. Hence
must incident on mirror normally.
Case 1: Object will be at focus of lens
$${1 \over f} = \left( {\mu - 1} \right)\left( {{1 \over R} - {1 \over { - R}}} \right) = {1 \over { - 18}}$$ $$ \Rightarrow $$ R = 18 cm
Case 2: Retraction at 1st surface:
$${1 \over { - 27}} - {{1.5} \over {{V_1}}} = {{1 - 1.5} \over R}$$ ... (i)
2nd retraction:
$${{1.5} \over {{V_1}}} - {\mu \over \infty } = {{15 - u} \over { - R}}$$ ...(ii)
From (i) and (ii)
$$\mu = {4 \over 3}$$
Case 1: Object will be at focus of lens
$${1 \over f} = \left( {\mu - 1} \right)\left( {{1 \over R} - {1 \over { - R}}} \right) = {1 \over { - 18}}$$ $$ \Rightarrow $$ R = 18 cm
Case 2: Retraction at 1st surface:
$${1 \over { - 27}} - {{1.5} \over {{V_1}}} = {{1 - 1.5} \over R}$$ ... (i)
2nd retraction:
$${{1.5} \over {{V_1}}} - {\mu \over \infty } = {{15 - u} \over { - R}}$$ ...(ii)
From (i) and (ii)
$$\mu = {4 \over 3}$$
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