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

(A) For a concave mirror,the mirror formula is $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.Using sign convention,$u$ is replaced by $-u$ and $f$ is repl

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) For a concave mirror,the mirror formula is $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.Using sign convention,$u$ is replaced by $-u$ and $f$ is replaced by $-f$.Thus,$\frac{1}{v} - \frac{1}{u} = -\frac{1}{f}$,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 far end of the rod is at infinity $(u = \infty)$,so its image is formed at the focus $(v = f)$.The near end of the rod is at distance $u$,so its image is at distance $v = \frac{uf}{u - f}$ from the pole.The length of the image is the difference between the positions of the two images: $L = |v - f|$.$L = |\frac{uf}{u - f} - f| = |\frac{uf - f(u - f)}{u - f}| = |\frac{uf - uf + f^2}{u - f}| = \frac{f^2}{u - f}$.

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

$A$ particle of charge '$q$' and mass '$m$' moves in a circular orbit of radius '$r$' with angular speed '$\omega$'. The ratio of the magnitude of its magnetic moment to that of its angular momentum depends on

If $A = \{1, 2, 3\}$ and $B = \{1, 4, 6, 9\}$,and $R$ is a relation from $A$ to $B$ defined by '$x$ is greater than $y$'. The range of $R$ is

The logically equivalent proposition of $p \Leftrightarrow q$ is

If $A+C=2B$,then $\frac{\cos C-\cos A}{\sin A-\sin C}$ is equal to

Two cards are drawn successively with replacement from a well-shuffled pack of $52$ cards. The mean of the number of kings is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module