A convex lens of focal length 12 cm is made | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The focal length of a lens in a medium is given by the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.For a lens i

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(D) The focal length of a lens in a medium is given by the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.For a lens in air $(f_a = 12 \ cm)$: $\frac{1}{f_a} = (\mu_g - 1)K$,where $K = (\frac{1}{R_1} - \frac{1}{R_2})$.For a lens in liquid $(f_l)$: $\frac{1}{f_l} = (\frac{\mu_g}{\mu_l} - 1)K$.Dividing the two equations: $\frac{f_l}{f_a} = \frac{\mu_g - 1}{\frac{\mu_g}{\mu_l} - 1}$.Substituting the values: $\frac{f_l}{12} = \frac{\frac{3}{2} - 1}{\frac{3/2}{5/4} - 1} = \frac{1/2}{\frac{6}{5} - 1} = \frac{1/2}{1/5} = \frac{5}{2}$.Therefore,$f_l = 12 	imes \frac{5}{2} = 30 \ cm$.

$A$ convex lens of focal length $12 \ cm$ is made of glass of refractive index $\mu_g = \frac{3}{2}$. What will be its focal length when immersed in a liquid of refractive index $\mu_l = \frac{5}{4}$? (in $cm$)

Similar Questions

Which of the following measurements require 'index correction'?

$A$ hollow convex lens of glass will behave like a

The figure shows a thin lens with centres of curvature $C_1$ and $C_2$. Find its focal length in $cm$. (Take refractive index $\mu = 1.5$)

$A$ real object approaches a fixed diverging lens with a constant velocity from infinity along the principal axis. The relative velocity between the object and its image will be

$A$ convex lens is dipped in a liquid whose refractive index is equal to the refractive index of the lens. Then its focal length will

Vedclass Test Series

Exam Paper Generator

Online Exam Module