Figure shows an object 'O' placed in front of a biconvex lens ( μ = 2 3 ) behind which there is water ( μ = 3 4 ) and a plane mirror at a distance of 20 m . A point object O is placed at

Question

Figure shows an object 'O' placed in front of a biconvex lens ( μ = 2 3 ​ ) behind which there is water ( μ = 3 4 ​ ) and a plane mirror at a distance of 20 m . A point object O is placed at a distance of 45 cm from the lens as shown in the figure. The position of the image of the object coincides with the object itself, then radius of curvature of the lens is :

TestHub · Accepted Answer

For the image to coincide with the object, rays must strike the plane mirror normally, i.e. they travel parallel inside the water.

First surface (air to glass): $\frac{3}{2 v _{1}} + \frac{1}{45} = \frac{3/2 - 1}{R} = \frac{1}{2 R}$ .

Second surface (glass to water, parallel output): $- \frac{3}{2 v _{1}} = \frac{4/3 - 3/2}{- R} = \frac{1}{6 R}$ .

Adding: $\frac{1}{45} = \frac{1}{2 R} + \frac{1}{6 R} = \frac{2}{3 R}$ ⇒ $R = 30$ cm - (D)

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