Question

JEE mains Previous year question
chapter :- Optics​

Answer 1

Answer:

$\sf Option\: b) 30 \:cm$

Solution:

To determine the radius of curvature of the curved surface, we can write

$\sf{R}^{2} = \sf {3}^{2} +( R - \frac{3}{10} )^{2}$

Solving this, we get

$\sf R = 15.15 \: cm$

Speed of light in any medium is $\sf \frac{c}{μ}$

where $\sf \: c = 3 \times {10}^{8} m {s}^{ - 1}$ and μ is the refractive index.

For the lens,

$\sf \: \frac{c}{μ} = 2 \times {10}^{8} m {s}^{ - 1} \\ \sfμ = \frac{3 \times {10}^{8} }{2 \times {10}^{8} } = 1.5$

Putting these values in the lens makers formula for plano convex lens, we have

$\sf \frac{1}{f} = (μ - 1)( \frac{1}{r} ) = (1.5 - 1)( \frac{1}{15.15} )c {m}^{ - 1}$

Solving this we get $\sf f = 30 \: cm$

$\tiny\sf The \: lens \: is \: as \: shown \: in \: the \: figure,$