The two surfaces of a concave lens, made of g | vedclass

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

Question

(A) Using the lens maker's formula:$\frac{1}{f} = \left( \frac{\mu_g}{\mu_m} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$Given that the r

Vedclass · Accepted Answer

becomes a convergent lens of focal length $3.5 R$

Vedclass · Answer

(A) Using the lens maker's formula:$\frac{1}{f} = \left( \frac{\mu_g}{\mu_m} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$Given that the refractive index of the glass lens is $\mu_g = 1.5$ and the refractive index of the surrounding medium is $\mu_m = 1.75$.For a biconcave lens, the radii of curvature are $R_1 = -R$ and $R_2 = +R$.Substituting these values into the formula:$\frac{1}{f} = \left( \frac{1.5}{1.75} - 1 \right) \left( \frac{1}{-R} - \frac{1}{R} \right)$$\frac{1}{f} = \left( \frac{1.5 - 1.75}{1.75} \right) \left( -\frac{2}{R} \right)$$\frac{1}{f} = \left( \frac{-0.25}{1.75} \right) \left( -\frac{2}{R} \right)$$\frac{1}{f} = \left( -\frac{1}{7} \right) \left( -\frac{2}{R} \right) = \frac{2}{7R}$$f = +3.5 R$Since the focal length $f$ is positive, the lens behaves as a convergent lens.

The two surfaces of a concave lens, made of glass of refractive index $1.5$, have the same radii of curvature $R$. It is now immersed in a medium of refractive index $1.75$. Then the lens:

Similar Questions

The diameter of a plano-convex lens is $6 \, cm$ and its thickness at the centre is $3 \, mm$. If the speed of light in the material of the lens is $2 \times 10^8 \, m/s$,the focal length of the lens is.......$cm$.

$A$ lens is made of flint glass (refractive index $= 1.5$). When the lens is immersed in a liquid of refractive index $1.25$,the focal length

$A$ thin lens made of glass (refractive index $= 1.5$) of focal length $f = 16 \; cm$ is immersed in a liquid of refractive index $1.42$. If its focal length in liquid is $f_{l}$,then the ratio $f_{l} / f$ is closest to the integer:

$A$ converging lens forms an image of an object on a screen. The image is real and twice the size of the object. If the positions of the screen and the object are interchanged,leaving the lens in the original position,the new image size on the screen is

Vedclass Test Series

Exam Paper Generator

Online Exam Module