The focal length of a convex lens of R.I. 1.5 is f when it is placed in air. When it is immersed in a liquid its focal length becomes xf. The refractive index of the liquid
Answer & explanation
Correct answer: option 2
$1/f=(n-1)(\frac{1}{R_1}+\frac{1}{R_2})$
$⇒1/f=(\frac{1.5}{1}-1)(\frac{1}{R_1}+\frac{1}{R_2})$ when the lens is placed in air and
$1/xf=(\frac{1.5}{y}-1)(\frac{1}{R_1}+\frac{1}{R_2})$ when the lens is places in the liquid.
Where y = R.I. of the liquid
$⇒\frac{f}{xf}=\frac{\frac{1.5}{y}-1}{1.5-1}$
$⇒1+\frac{1}{2x}=\frac{1.5}{y}$
$⇒y=\frac{3}{2}(\frac{2x}{2x+1})$
$⇒y=\frac{3x}{2x+1}$