Question

JEE mains Previous year question
chapter :- Geometrical Optics​

Answer 1

Answer in the attachment.

[refractive index of liquid medium was not given in the question so I've mentioned it.]

Answer 2

$\Large{\underline{\underline{\bold{✯Given:}}}}$

• Optical Power of thin glass lens = -5D.

$\Large{\underline{\underline{\red{\bold{➸\;To\; Find:}}}}}$

• Optical Power of the thin glass lens in a liquid medium.

$\Large{\underline{\underline{\bold{\color{salmon}๛\; Required\; Response:}}}}$

Refractive Index of the thin glass lens $\bf{\mu_g = 1.5}$

We Know That

• $\Large\pink{\bf{\frac{1}{f} = (n-1)(\frac{1}{R_1}-\frac{1}{R_2})}}$

$→\;\Large{\frac{1}{f_a} = (\frac{1.5}{1}-1)(\frac{1}{R_1}-\frac{1}{R_2})}$ ____①

$→\;\Large{\frac{1}{f_m} = (\frac{μ_g}{μ_m}-1)(\frac{1}{R_1}-\frac{1}{R_2})}$____②

Divide ① by ②

$\Large\frac{f_a}{f_m} = [\frac{1.5-1}{\frac{1.5}{1.6}-1}] = -8$

As, $\bold{P_a = -5}$

• $\pink{\bf{p = \frac{1}{f}}}$

$\Large{P_a = \frac{\mu}{f_a}}$

• $\pink{\bf{μ_0 = 4\pi×10^{-7} = 1.25 = 1}}$

$P_a = \frac{1}{f_a}$

$→\;\Large{-5 = \frac{1}{f_a}}$

$→\;\Large{f_a = -\frac{1}{5}}$___③

$→\;\Large{\sf{\frac{f_m}{f_a} = -8}}$

$→\;\Large{f_m = -8×f_a}$

$→\;\Large{f_m = -8×\frac{1}{-5}}$ (From ③)

$→\Large{\;f_m = \frac{\cancel{-}8×1}{\cancel{-}5}}$

$→\;\Large{f_m = \frac{8}{5}}$

As, $\bold{P_m = \frac{\mu}{f_m}}$

$→\;\Large{P_m = \frac{1.6}{8}×5}$

$→\;\Large{P_m = 0.2×5}$

$➝\;\huge{\sf{P_m = 1}}$

$\Large{\bold{Option\;-\;A}}\;→\;{\bold{\orange{1D}}}$

• ᴏᴘᴛɪᴄᴀʟ ᴘᴏᴡᴇʀ ᴏꜰ ᴛʜᴇ ᴛʜɪɴ ɢʟᴀꜱꜱ ʟᴇɴꜱ ɪɴ ᴀ ʟɪQᴜɪᴅ ᴍᴇᴅɪᴜᴍ ☞ $\huge{\blue{\bf{1ᴅ}}}$

$\boxed{\tt{@MrSoverign}}$

Hope This Helps!!