When the object is at distances u_1 and u_2 f | vedclass

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

Question

(B) Let the focal length of the lens be $f$. The magnification $m$ is given by $m = \frac{v}{u}$.For a real image at distance $u_1$,the magnification

Vedclass · Accepted Answer

$\frac{u_1 + u_2}{2}$

Vedclass · Answer

(B) Let the focal length of the lens be $f$. The magnification $m$ is given by $m = \frac{v}{u}$.For a real image at distance $u_1$,the magnification is $m_1 = \frac{f}{f - u_1}$. Since the image is real,$m_1$ is negative,so $m_1 = -\frac{f}{u_1 - f}$.For a virtual image at distance $u_2$,the magnification is $m_2 = \frac{f}{f - u_2}$. Since the image is virtual,$m_2$ is positive,so $m_2 = \frac{f}{f - u_2}$.Given that the sizes are the same,$|m_1| = |m_2|$.Thus,$|-\frac{f}{u_1 - f}| = |\frac{f}{f - u_2}|$.Since $u_1 > f$ (for real image) and $u_2 This simplifies to $f - u_2 = u_1 - f$.Rearranging the terms,we get $2f = u_1 + u_2$,which implies $f = \frac{u_1 + u_2}{2}$.

When the object is at distances $u_1$ and $u_2$ from a lens,the images formed are real and virtual respectively,and both have the same size. Then the focal length of the lens is:

Similar Questions

$A$ lamp is placed $6.0\; m$ from a wall. On putting a lens between the lamp and the wall at a distance of $1.2\; m$ from the lamp,a real image of the lamp is formed on the wall. The magnification of the image is

$A$ convex lens of focal length $25 \ cm$ and made of glass with refractive index $1.5$ is immersed in water. The absolute change in focal length of the glass is [Use refractive index of water = $\frac{4}{3}$] (in $cm$)

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

The relation between $n_1$ and $n_2$,if the behavior of light rays is as shown in the figure,is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module