Question: Q. 5. The radius of curvature of the curved surface of a planoconvex lens is $20 \mathrm{~cm}$. If the refractive index of the material of the lens be 1.5 , it will
(a) act as a convex lens only for the objects that lie on its curved side.
(b) act as a concave lens for the objects that lie on its curved side.
(c) act as a convex lens irrespective of the side on which the object lies.
(d) act as a concave lens irrespective of side on which the object lies. [NCERT Exemplar]
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Solution:
Ans. Correct option : (c)
Explanation: As we know the relations between $f, \mu, R_{1}$ and $R_{2}$ is known as lens maker’s formula :
$$ \begin{aligned} & \frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \ & R_{1}=\infty, R_{2}=-R \ & f=\frac{R}{(\mu-1)} \end{aligned} $$
Given that,
$$ \begin{aligned} R & =20 \mathrm{~cm}, \ \mu & =1.5 \ f & =\frac{R}{\mu-1}=\frac{20}{15-1}=40 \mathrm{~cm} \end{aligned} $$
Put the values;
As $f>0$, it means converging nature of the lens.
So, lens act as a convex lens irrespective of the side on which the object lies.
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