Question

A biconvex lens of glass of refractive index 1.5 having focal length of 20 cm is placed in a medium of refractive index 1.65. Find its focal length. What should be the value of refractive index of the medium in which the lens should be placed so that it acts as a plane sheet of glass

Answer 1

focal length will increase​ 22cm.

Answer 2

According to the data given

Refractive index of biconvex glass (n) = 1.5

Focal length = 20cm

Refractive index of the medium in which the biconvex lens is placed ($n_{m}$)= 1.65

The focal length of a biconvex lens is given by

$\frac{1}{f} =(n-1) (\frac{1}{R_{1} } -\frac{1}{R_{2} } )$

where f = focal length of the lens

n= Refractive index of the glass lens

$(\frac{1}{R_{1} } -\frac{1}{R_{2} } )$ = Radius of the biconvex lens

∴ $\frac{1}{20} = (1.5-1)(\frac{1}{R_{1} } -\frac{1}{R_{2} } )$

⇒ $\frac{1}{20} = 0.5(\frac{1}{R_{1} } -\frac{1}{R_{2} } )$ ...... Equation (1)

Focal length of biconvex lens in the medium of refractive index is 1.65 is

$\frac{1}{f_{1} } = (1.65-1)(\frac{1}{R_{1} } -\frac{1}{R_{2} } )$

∴$\frac{1}{f_{1} } = (0.65)(\frac{1}{R_{1} } -\frac{1}{R_{2} } )$ ........Equation (2)

Dividing Equation(1) by Equation(2) we get

$\frac{f_{1} }{20} = \frac{0.5}{0.65}$

∴ f₁ = 15.38 cm

Hence focal length of the biconvex lens in the medium of refractive index of 1.65 is 15.38 cm.

For the second part of the question

Focal length of a plane glass is (f) = ∞

∴$\frac{1}{f} = 0$ ...... Equation(3)

Substituting the value of equation 3 in the below mentioned equation of focal length of a biconvex lens

$\frac{1}{f} =(n-1) (\frac{1}{R_{1} } -\frac{1}{R_{2} } )$

∴ $0 =(n-1) (\frac{1}{R_{1} } -\frac{1}{R_{2} } )$

⇒ n-1=0

⇒ n= 1

Hence refractive index of the plane sheet of glass is 1 .