Question

A thin lens of refractive index 1.5 has a focal length of 15 cm in air. When the lens is placed in a medium of refractive index 4/3, its focal length will become :
(a) 30 cm
(b) 60 cm
(c) 45 cm
(d) 75 cm​

Answer 1

Answer: it will be B 60

Explanation:

Answer 2

Answer:

The focal length when lens is placed in a medium of refractive index 4/3 is 60 cm.

Explanation:

Given that :

• Refractive index of lens = 1.5
• Focal length of lens = 15 cm
• Refractive index of medium = 4/3

Solution :

• Let, n1 be the refractive index of air; n2 be the refractive index of lens and n3 be the refractive index of medium.
• From lens maker's formula

$\frac{1}{f} = ( \frac{n2}{n1} - 1)( \frac{1}{R1} - \frac{1}{R2} )$

where, n2 is refractive index of second medium and n1 is refractive index of first medium.

R1 and R2 are radius of curvatures of lens.

• For air and lens system :

$\frac{1}{15} = ( \frac{1.5}{1} - 1)( \frac{1}{R1} - \frac{1}{R2} ) \\( \frac{1}{R1} - \frac{1}{R2} ) = \frac{1}{15 \times 0.5} \\ (\frac{1}{R1} - \frac{1}{R2} ) = \frac{1}{7.5} - - - (1)$

• For medium and lens system, the new focal length be f1 :

$\frac{1}{f1} = ( \frac{n2}{n3} - 1)(\frac{1}{R1} - \frac{1}{R2} ) \\ \frac{1}{f1} = ( \frac{1.5 \times 3}{4} - 1)(\frac{1}{R1} - \frac{1}{R2} ) \\ \frac{1}{f1} = ( \frac{0.5)}{4} (\frac{1}{R1} - \frac{1}{R2}) \\ using \: value \:from \: equation \: (1) \\ \frac{1}{f1} = \frac{0.5}{4} \times \frac{1}{7.5} \\ \frac{1}{f1} = \frac{1}{60} \\ f = 60 \: cm$

• Hence, the new focal length in medium is 60 cm .