An infinitely long rod lies along the axis of | vedclass

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

Question

(D) For the near end of the rod,the mirror formula is $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$.Using the sign convention for a concave mirror,$u$ and

Vedclass · Accepted Answer

$\frac{f^2}{u - f}$

Vedclass · Answer

(D) For the near end of the rod,the mirror formula is $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$.Using the sign convention for a concave mirror,$u$ and $f$ are negative.Thus,$\frac{1}{-f} = \frac{1}{-u} + \frac{1}{v}$,which gives $\frac{1}{v} = \frac{1}{u} - \frac{1}{f} = \frac{f - u}{uf}$.So,$v = \frac{uf}{f - u} = -\frac{uf}{u - f}$. The distance of the image from the mirror is $|v| = \frac{uf}{u - f}$.The far end of the rod is at infinity $(u = \infty)$. According to the mirror formula,the image of the far end is formed at the focus $(v = f)$.The length of the image is the difference between the image positions of the two ends: $L = |v| - f$.$L = \frac{uf}{u - f} - f = \frac{uf - f(u - f)}{u - f} = \frac{uf - uf + f^2}{u - f} = \frac{f^2}{u - f}$.

An infinitely long rod lies along the axis of a concave mirror of focal length $f$. The near end of the rod is at a distance $u > f$ from the mirror. Its image will have a length

Similar Questions

If $\int e^x \left( \frac{1-\sin x}{1-\cos x} \right) dx = f(x) + \text{constant}$,then $f(x)$ is equal to

The line $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z + 1}{4}$ meets the plane $x + 2y + 3z = 14$ at the point:

The vapour pressure in $mm$ of $Hg$,of an aqueous solution obtained by adding $18 \ g$ of glucose $(C_6H_{12}O_6)$ to $180 \ g$ of water at $100^{\circ} C$ is

Curie temperature is the temperature above which :

The orbital angular momentum of a satellite revolving at a distance $r$ from the centre is $L$. If the distance is increased to $4r$,then the new angular momentum will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module