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Question

As shown in the figure, region BCDEF and ABFG are of refractive index 2.0 and 1.5 respectively. A particle O is kept at the mid-point of DH. Image of the object as seen by the eye is at a distance of :


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Solution

Here the object is placed in medium (μ=2),
Thus, μ1=2
Observer is in medium, μ2=1

Sign convention: Direction of incident ray is taken as +ve.
(ray emerging from O and moving toward D is positive.)

By using the equation :

μ2vμ1u=μ2μ1R

Here, u=202=10 cm

R=10 cm

1v2(10)=1210

or, 1v+210=110

or, 1v=110210=110

v=10 cm

Thus, image is formed at 10 cm from point D.

Why this question?
Bottom line: The observer (in medium μ2=1) will see the image of object (in medium μ1=2), due to refraction from spherical surface CDE.

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