Geometrical OpticsHard

Question

A biconvex lens is formed by using two thin planoconvex lenses, as shown in the figure. The refractive index and radius of curved surfaces are also mentioned in figure. When an object is placed on the left side of lens at a distance of 30 cm from the biconvex lens, the magnification of the image will be :

Options

A.-2
B.+2
C.+2.5
D.-2.5

Solution

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f_{net}} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$

$${\frac{1}{v} + \frac{1}{30} = (1.5 - 1)\left( \frac{1}{15} - \frac{1}{\infty} \right) + (1.2 - 1)\left( \frac{1}{\infty} + \frac{1}{12} \right) }{\frac{1}{v} + \frac{1}{30} = \frac{1}{30} + \frac{1}{60} }{v = 60 }{m = \frac{v}{u} = \frac{60}{- 30} = - 2}$$

Create a free account to view solution

View Solution Free
Topic: Geometrical Optics·Practice all Geometrical Optics questions

More Geometrical Optics Questions