A thin converging lens of focal length f=25 | vedclass

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

Question

(C) According to the first condition:$f = 25 \ cm, v = 75 \ cm$Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:$\frac{1}{25} = \frac{

Vedclass · Accepted Answer

$12.5$

Vedclass · Answer

(C) According to the first condition:$f = 25 \ cm, v = 75 \ cm$Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:$\frac{1}{25} = \frac{1}{75} - \frac{1}{u}$$\frac{1}{u} = \frac{1}{75} - \frac{1}{25} = \frac{1-3}{75} = -\frac{2}{75}$$u = -37.5 \ cm$According to the second condition,the screen is moved closer by $25 \ cm$,so the new image distance is $v_1 = 75 - 25 = 50 \ cm$.Using the lens formula again:$\frac{1}{25} = \frac{1}{50} - \frac{1}{u_1}$$\frac{1}{u_1} = \frac{1}{50} - \frac{1}{25} = \frac{1-2}{50} = -\frac{1}{50}$$u_1 = -50 \ cm$The distance through which the object has to be shifted is:$\Delta u = |u_1| - |u| = 50 \ cm - 37.5 \ cm = 12.5 \ cm$.

Similar Questions

Explain the structures of diborane and boric acid.

The method of asexual reproduction in algae is -

$A$ frictionless track $ABCDE$ ends in a circular loop of radius $R$. $A$ body slides down the track from point $A$ which is at a height $h = 5 \, cm$. The maximum value of $R$ for the body to successfully complete the loop is .................. $cm$.

How many glucose molecules are required for the formation of $52$ pyruvic acid molecules at the end of glycolysis?

The number of possible straight lines passing through $(2, 3)$ and forming a triangle with the coordinate axes,whose area is $12 \, sq. \, units$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module