A convex lens of refractive index 1.5 and foc | vedclass

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

Question

(C) Using the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.In air: $\frac{1}{f_{	ext{air}}} = (1.5 -

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) Using the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.In air: $\frac{1}{f_{	ext{air}}} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight) = 0.5 	imes K$, where $K = (\frac{1}{R_1} - \frac{1}{R_2})$.Given $f_{	ext{air}} = f = 18 \ cm$, so $K = \frac{1}{0.5f} = \frac{2}{f}$.In water $(\mu_w = 4/3)$: $\frac{1}{f_{	ext{water}}} = (\frac{1.5}{4/3} - 1) K = (\frac{4.5}{4} - 1) K = (1.125 - 1) K = 0.125 K$.Substituting $K = \frac{2}{f}$: $\frac{1}{f_{	ext{water}}} = 0.125 	imes \frac{2}{f} = \frac{0.25}{f} = \frac{1}{4f}$.Thus, $f_{	ext{water}} = 4f$.The difference in focal lengths is $f_{	ext{water}} - f_{	ext{air}} = 4f - f = 3f$.Comparing with $\alpha 	imes f$, we get $\alpha = 3$.

$A$ convex lens of refractive index $1.5$ and focal length $f = 18 \ cm$ is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is $\alpha \times f$. The value of $\alpha$ is . . . . . . . (Refractive index of water $= 4/3$)

Similar Questions

The dispersive powers of crown and flint glasses are $0.02$ and $0.04$ respectively. In an achromatic combination of lenses,the focal length of the flint glass lens is $40 \ cm$. The focal length of the crown glass lens will be $....... \ cm$.

The radius of curvature of a convex lens is $40 \,cm$ for each surface. Its refractive index is $1.5$. Its focal length is: (in $\,cm$)

Write the Gaussian equation for a lens.

$A$ lens of refractive index $n$ is placed in a liquid of refractive index $n'$. If the focal length of the lens in air is $f$,what will be its focal length in the liquid?

The dispersive powers of the glasses of lenses used in an achromatic pair are in the ratio $5 : 3$. If the focal length of the concave lens is $15 \ cm$,then the nature and focal length of the other lens would be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module