Find the magnification for the lens. The foca | vedclass

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

Question

(B) From the given image,the object distance is $u = -20 \; cm$ and the focal length of the convex lens is $f = +25 \; cm$.Using the lens formula:$\fr

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) From the given image,the object distance is $u = -20 \; cm$ and the focal length of the convex lens is $f = +25 \; cm$.Using the lens formula:$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$Substituting the values:$\frac{1}{v} - \frac{1}{-20} = \frac{1}{25}$$\frac{1}{v} + \frac{1}{20} = \frac{1}{25}$$\frac{1}{v} = \frac{1}{25} - \frac{1}{20}$$\frac{1}{v} = \frac{4 - 5}{100}$$\frac{1}{v} = -\frac{1}{100}$$v = -100 \; cm$The magnification $m$ is given by:$m = \frac{v}{u}$$m = \frac{-100}{-20}$$m = 5$

Find the magnification for the lens. The focal length is $25 \; cm$.

Similar Questions

$A$ concave-convex lens of refractive index $1.5$ has radii of curvature of its surfaces as $30 \ cm$ and $20 \ cm$,respectively. The concave surface is facing upwards and is filled with a liquid of refractive index $1.3$. The focal length of the liquid-glass combination will be

$A$ convex lens produces a real image $m$ times the size of the object. What will be the distance of the object from the lens?

The radius of curvature for a convex lens is $40 \ cm$ for each surface. Its refractive index is $1.5$. The focal length will be ...... $cm$.

If the central portion of a convex lens is wrapped in black paper as shown in the figure,

An achromatic telescope objective is to be made by combining the lenses of flint and crown glasses. This proper choice is

Vedclass Test Series

Exam Paper Generator

Online Exam Module