1. Tardigrade
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  3. Physics
  4. A pin is placed 10 cm in front of a convex lens of focal length 20 cm and made of a material of refractive index 1.5. The convex surface of the lens farther away from Jhe pin is silvered and has a radius of curvature of 22 cm. Determine the position of the final image. Is the image real or virtual ?

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Q. A pin is placed 10 cm in front of a convex lens of focal length 20 cm and made of a material of refractive index 1.5. The convex surface of the lens farther away from Jhe pin is silvered and has a radius of curvature of 22 cm. Determine the position of the final image. Is the image real or virtual ?

IIT JEEIIT JEE 1978

Solution:

The given system behaves like a mirror of power given by $P = 2P_L + P_M$ $or \, \, \, \, -\frac{1}{F}=2\Big(\frac{1}{0.2}\Big)+\Big(\frac{-2}{0.22}\Big)$ As $P_L=\frac{1}{f(m)}$ and $P_M=\frac{-1}{f(m)}=frac{-2}{R(m)}$ Solving this equation, we get $ $ F = -1.1 m = - 110cm i.e. the system behaves as a concave mirror of focal length 18.33 cm Using the mirror formula $ \frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ we have $ \frac{1}{v}-\frac{1}{10}=-\frac{1}{110}$ or $ c=11 cm$ i.e. virtual image will be formed at a distance of 11 cm.

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