Ray Optics and Optical Instruments - Result Question 41
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42. The radius of curvature of a thin plano-convex lens is $10 cm$ (of curved surface) and the refractive index is 1.5. If the plane surface is silvered, then it behaves like a concave mirror of focal length
======= ####42. The radius of curvature of a thin plano-convex lens is $10 cm$ (of curved surface) and the refractive index is 1.5. If the plane surface is silvered, then it behaves like a concave mirror of focal length
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/ray-optics-and-optical-instruments/ray-optics-and-optical-instruments—result-question-41.md (a) $10 cm$
(b) $15 cm$
(c) $20 cm$
(d) $5 cm$
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Answer:
Correct Answer: 42. (a)
Solution:
- (a) The silvered plano convex lens behaves as a concave mirro; whose focal length is given by $\frac{1}{F}=\frac{2}{f_1}+\frac{1}{f_m}$
If plane surface is silvered
$f_m=\frac{R_2}{2}=\frac{\infty}{2}=\infty$
$\therefore \frac{1}{f_1}=(\mu-1)(\frac{1}{R_1}-\frac{1}{R_2})$
$=(\mu-1)(\frac{1}{R}-\frac{1}{\infty})=\frac{\mu-1}{R}$
$\therefore \frac{1}{F}=\frac{2(\mu-1)}{R}+\frac{1}{\infty}=\frac{2(\mu-1)}{R}$
$F=\frac{R}{2(\mu-1)}$
Here, $R=10 cm, \mu=1.5$
$\therefore F=\frac{10}{2(1.5-1)}=10 cm$
43 .
(b) $R_1=60 cm, R_2=\infty, \mu=1.6$
$\frac{1}{f}=(\mu-1)(\frac{1}{R_1}-\frac{1}{R_2})$
$\frac{1}{f}=(1.6-1)(\frac{1}{60}) \Rightarrow f=100 cm$.
