Optics Question 99
Question: A spherical surface of radius of curvature R separates air (refractive index 1.0) from glass (refractive index 1.5). The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at a point O, and PO = OQ. The distance PO is equal to
[IIT JEE 1998; DPMT 2000]
Options:
A) 5 R
B) 3 R
C) 2 R
D) 1.5 R
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{{\mu _{2}}}{v}-\frac{{\mu _{1}}}{u}=\frac{{\mu _{2}}-{\mu _{1}}}{R}\Rightarrow \frac{1.5}{+OQ}-\frac{1}{(-OP)}=\frac{(1.5-1)}{+R} $ On putting $ OQ=OP,OP=5R $
