Question

A concave lens of refractive index 1.5 has both surfaces of the same radius of curvature R. On immersion in a medium of refractive index 1.75, it will behave as a

• A. Converging lens of focal length 3.5R
• B. Converging lens of focal length 3R
• C. Diverging lens of focal length 3.5R
• D. Diverging lens of focal length 3R

Answer 1

The correct option is A Converging lens of focal length 3.5R

$\frac{1}{f}=\frac{1}{\text{focal lengthof the lens}}=(\frac{{\mu }_2}{{\mu }_1}-1)(\frac{1}{R1}-\frac{1}{R2})Where{\mu }_2=\text{Refractive index of lens = 1.75}{\mu }_1=\text{Refractive index of surroundings = 1.75}\text{Using sign conventions, distance}\text{measured in the direction of the}\text{incident ray is positive and in the}\text{opposite direction is negative.}\text{It is biconvex lens hence both surfaces}\text{have the same radius of curvature}\text{let's say R}R1=-RR2=RBySolving,f=3.5R\text{Focal length is positive, hence it is}\text{converging lens.}$