A thin lens made of glass (refractive index = | vedclass

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

Question

(C) Using the Lens Maker's Formula: $\frac{1}{f} = (\mu_{g} - 1) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)$.For the lens in air $(\mu_{a} = 1)$

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) Using the Lens Maker's Formula: $\frac{1}{f} = (\mu_{g} - 1) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} ight)$.For the lens in air $(\mu_{a} = 1)$:$\frac{1}{f} = (1.5 - 1) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} ight) = 0.5 \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} ight) \dots (1)$For the lens in liquid $(\mu_{l} = 1.42)$:$\frac{1}{f_{l}} = \left( \frac{1.5}{1.42} - 1 ight) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} ight) = \left( \frac{1.5 - 1.42}{1.42} ight) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} ight) = \frac{0.08}{1.42} \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} ight) \dots (2)$Dividing equation $(1)$ by equation $(2)$:$\frac{f_{l}}{f} = \frac{0.5 \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} ight)}{\frac{0.08}{1.42} \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} ight)} = \frac{0.5 	imes 1.42}{0.08} = \frac{0.71}{0.08} = 8.875$.The value $8.875$ is closest to the integer $9$.

$A$ 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:

Similar Questions

An object of height $1.5 \ cm$ is placed on the axis of a convex lens of focal length $25 \ cm$. $A$ real image is formed at a distance of $75 \ cm$ from the lens. The size of the image will be........$cm$.

$A$ plano-convex lens is made of a material with a refractive index of $1.6$. The radius of curvature of the curved surface is $60 \ cm$. The focal length of the lens is.....$cm$.

An object is located in a fixed position in front of a screen. $A$ sharp image is obtained on the screen for two positions of a thin lens separated by $10\, cm$. The size of the images in the two situations are in the ratio $3 : 2$. What is the distance between the screen and the object in $cm$?

When monochromatic red light is used instead of blue light in a convex lens, its focal length will

When an object is at a distance $u_1$ and $u_2$ from a lens,a real image and a virtual image are formed,respectively,having the same magnification $m$. The focal length of the lens is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module