Question 159: Ray Optics

Question

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 μ
1
will be :-
![](https://imagex.cdn.examgoal.net/3rsa0w2w9jwotowyj/c707f11d-4eb0-4a89-aca0-d160da6ed00d/e40313a0-8aa3-11e9-86c1-73f7f991b766/file-3rsa0w2w9jwotowyk.png)

Accepted Answer

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 1
st
surface:
${1 \over { - 27}} - {{1.5} \over {{V_1}}} = {{1 - 1.5} \over R}$    ... (i)
2
nd
retraction:
${{1.5} \over {{V_1}}} - {\mu  \over \infty } = {{15 - u} \over { - R}}$   ...(ii)
From (i) and (ii)
$\mu  = {4 \over 3}$

Answer

$\sqrt 3 $

Answer

3 / 2

Answer

$\sqrt 2 $

Answer

4 / 3

Ray Optics

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