A small linear object of length b is placed o | vedclass

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

Question

(D) From the mirror formula,$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.Differentiating both sides with respect to $u$,we get $-\frac{1}{v^2} dv - \frac

Vedclass · Accepted Answer

$b \left( \frac{f}{u - f} \right)^2$

Vedclass · Answer

(D) From the mirror formula,$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.Differentiating both sides with respect to $u$,we get $-\frac{1}{v^2} dv - \frac{1}{u^2} du = 0$.This implies $dv = -\frac{v^2}{u^2} du$. The magnitude of the image length $|dv|$ is given by $|dv| = \left( \frac{v}{u} \right)^2 |du|$.Since the object is small,$|du| = b$,so the image length is $L = \left( \frac{v}{u} \right)^2 b$.From the mirror formula,$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{u - f}{fu}$,which gives $\frac{v}{u} = \frac{f}{u - f}$.Substituting this into the expression for $L$,we get $L = b \left( \frac{f}{u - f} \right)^2$.

$A$ small linear object of length $b$ is placed on the axis of a concave mirror of focal length $f$ at a distance $u$ from its pole. What is the approximate length of the image?

Similar Questions

An object placed in front of a concave mirror of focal length $0.15 \,m$ produces a virtual image,which is twice the size of the object. The position of the object with respect to the mirror is ........... $cm$. (in $.5$)

An object of length $f/3$ is placed on the principal axis of a concave mirror of focal length $f$. The image of the end of the object farthest from the mirror is formed at the same position. Find the length of the image.

If the coordinates of the position of an object are $(-60 \, cm, 6 \, cm)$,then the coordinates of the image will be:

While capturing solar energy for commercial purposes,we use:

If an object is placed $10 \ cm$ in front of a concave mirror of focal length $20 \ cm,$ the image will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module