A thin lens of refractive index 1.5, and focal length in air 20 cm is placed inside, a large container containing two immiscible liquids as shown below. If an object is placed at an infinite distance close to principle axis (common intersection line of layers), the distance between images

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25 cm

45 cm

65 cm

85 cm

answer is C.

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Detailed Solution

For upper portion, $f_{1} = \frac{μ_{2} - 1}{\frac{μ_{2}}{μ_{1}} - 1} (f_{a}) = \frac{1.5 - 1}{\frac{1.5}{1.2} - 1} \times 20 = 40 c m$ $f_{a} = f o c a l l e n g t h o f t h e l e n s i n a i r .$ $f_{1} i s t h e p o s i t i o n o f t h e i m a g e f o r t h e u p p e r p a r t o f t h e o b j e c t a n d f_{2} i s t h e p o s i t i o n o f t h e i m a g e f o r t h e l o w e r p a r t o f t h e o b j e c t .$

For lower portion, $f_{1} = \frac{μ_{2} - 1}{μ_{2} / μ_{3} - 1} (f_{a}) = \frac{1.5 - 1}{1.5 / 2.5 - 1} \times 20 = - 25 c m$ so the distance between the images is equal to 40+25=65 cm.