Question

Focal length of a convex lens of refractive index 1.5 is 2cm. Focal length of lens when immersed in a liquid of refractive index 1.25 will be

• A. 5cm
• B. 7.5cm
• C. 10cm
• D. 2.5cm

Answer 1

The correct option is A 5cm

$\frac{1}{f_a}=\frac{1}{\text{focal length of the lens in air}}=({\mu }_g-1)(\frac{1}{R1}-\frac{1}{R2})=2cm\frac{1}{f_1}=\frac{1}{\text{focal length of the lens in liquid}}=(\frac{{\mu }_g}{{\mu }_l}-1)(\frac{1}{R1}-\frac{1}{R2})\frac{f_1}{f_a}=\frac{({\mu }_g-1)}{(\frac{{\mu }_g}{{\mu }_f}-1)}{\mu }_l=1.25and{\mu }_g=1.5\frac{f_1}{f_a}=\frac{5}{2}{f}_1=5cm$