One surface of a thin equniconvex lens of radii of curvature 20cm is made of glass of refractive index $\frac{3}{2}$ is silvered. In exp-I an small pin is kept on optic axis transversly at a distance 8cm from lens. In exp-II same object is kept in same manner as in exp-I, but whole system is immersed in large liquid medium of refractive index 2.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

In exp-I lens behaves as converging element while in experiment-II lens behaves as diverging element

Height of image in two experiment is same

In exp-I a real, inverted image is formed

In exp-II a virtual, inverted image is formed

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

exp-1
Question Image
$\frac{1}{- 8} + \frac{1}{v} = \frac{2 (\frac{3}{2} - 1)}{R} + \frac{2 (\frac{3}{2})}{R}$
$v = - \frac{8 \times 5}{8 - 5} = - \frac{40}{3}; \Rightarrow$ real, magnified inverted image.
exp-2
$\frac{μ_{1}}{- 8} + \frac{μ_{1}}{v} = \frac{2 (\frac{3}{2} - 2)}{R} + \frac{2 (\frac{3}{2})}{R}$
$v = - (\frac{8 \times 20}{8 - 20}) = \frac{40}{3} \Rightarrow$ virtual, magnified and erect image.