A thin plano-convex lens (μ=32) of focal length 12.5 cm is placed on a horizontal plane mirror as shown in the figure. The space between the lens and the mirror is filled with water (μw=43). Find the distance (in mm) above the lens at which a point object is placed on its principal axis so that the object coincides with its own image.

1f1=(32−1)(1∞+1R)=12R2R=f1⇒R=f12=254cm1f2=(43−1)(1−R−1∞)=−13Rf2=−3R=754cmNOW, −1F=2f1+2f2−1fm−1F=425−875−1∞⇒F=−754cmThe point object is placed at a distance, d=752=37.50cm

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A thin plano-convex lens $(μ = \frac{3}{2})$ of focal length 12.5 cm is placed on a horizontal plane mirror as shown in the figure. The space between the lens and the mirror is filled with water $(μ_{w} = \frac{4}{3}) .$ Find the distance (in mm) above the lens at which a point object is placed on its principal axis so that the object coincides with its own image.

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Detailed Solution

$\frac{1}{f_{1}} = (\frac{3}{2} - 1) (\frac{1}{\infty} + \frac{1}{R}) = \frac{1}{2 R} 2 R = f_{1} \Rightarrow R = \frac{f_{1}}{2} = \frac{25}{4} c m$

$\frac{1}{f_{2}} = (\frac{4}{3} - 1) (\frac{1}{- R} - \frac{1}{\infty}) = - \frac{1}{3 R} f_{2} = - 3 R = \frac{75}{4} c m$

NOW, $- \frac{1}{F} = \frac{2}{f_{1}} + \frac{2}{f_{2}} - \frac{1}{f_{m}} - \frac{1}{F} = \frac{4}{25} - \frac{8}{75} - \frac{1}{\infty} \Rightarrow F = - \frac{75}{4} c m$