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  • A plane convex lens of refractive index 1.5 and radius of curvature 30 cm is silvered at the curved surface. Now this lens has been used to form the image of an object. At what distance from t

A plane convex lens of refractive index 1.5 and radius of curvature 30 cm is silvered at the curved surface. Now this lens has been used to form the image of an object. At what distance from this lens an object be placed in order to have a real image of the size of the

Option: 1

20 cm


Option: 2

30 cm


Option: 3

60 cm


Option: 4

80 cm


Answers (1)

best_answer

A plano-convex lens behaves as a concave mirror, if its one surface (curved) is silvered. The rays refracted from plane surface are reflected from curved surface and again refract from plane surface. Therefore, in this lens two refractions and one reflection occur.

Let the focal length of silvered lens be F.

\begin{aligned} \frac{1}{F} &=\frac{1}{f}+\frac{1}{f}+\frac{1}{f_{m}} \\ &=\frac{2}{f}+\frac{1}{f_{m}} \end{aligned}

where, f = focal length of lens before silvering

fm = local length of spherical mirror. 

\\ \frac{1}{F}=\frac{2}{f}+\frac{2}{R} \quad\left[\because R=2 f_{m}\right] \ldots(\mathrm{i})\\ Now, \frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\ldots( ii)

\\ Here, R_{1}=\infty, R_{2}=30 \mathrm{cm}\\ \frac{1}{f}=(1.5-1)\left [ \frac{1}{\infty}-\frac{1}{30} \right ].\\ or \ \ \frac{1}{f}=-\frac{0.5}{30}=-\frac{1}{60} \\ or\ \ f=-60 \mathrm{cm}

\\ Hence, from\ \ Eq. (i), we \ \ get \\ \\ \begin{array}{l} \frac{1}{F}=\frac{2}{60}+\frac{2}{30}=\frac{6}{60} \\\\ F=10 \mathrm{cm} \end{array}

Again given that, Size of object = Size of image

i.e., \quad O=I \therefore m=-\frac{v}{u}=\frac{I}{O} \Rightarrow \quad \frac{v}{u}=-1 or \quad v=-u

\\ Thus,\ from \ lens\ formula,\\ \begin{aligned} \frac{1}{F}=\frac{1}{v}-\frac{1}{u} & \Rightarrow \frac{1}{10}=\frac{1}{-u}-\frac{1}{u} \\ \frac{1}{10} &=-\frac{2}{u} \\ u &=-20 \mathrm{cm} \end{aligned}

Hence, to get a real image, object must be placed at a distance 20 cm on the left side of lens.

Posted by

shivangi.shekhar

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