Question

An object starts moving at an angle at ${45}^o$ with the principal axis as shown in fig in front of a biconvex lens of focal length $+10cm$. If $\theta$ denotes the angle at which image starts to move with principal axis, then:

• A. $\theta =\frac{3\pi }{4}$
• B. $\theta =\frac{\pi }{2}$
• C. $\theta =\frac{\pi }{4}$
• D. $\theta =-\frac{\pi }{4}$

Answer 1

The correct option is D $\theta =-\frac{\pi }{4}$

$u=-20$ ; $f=+10$
So, from $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$, we get:
$v=+20$
Magnification: $\frac{v}{u}=-1$............ (i)
Hence vertical velocity of image is same as that of object but in reverse direction.
Let the horizontal velocity be $V_{Ix}$
We have: $v=\frac{fu}{u-f}$
Differentiating w.r.t. time, we get:
$V_{Ix}=\frac{f^2}{(u-f{)}^2}\times V_{object}$
$\Rightarrow$ $V_{Ix}=V_{object}$ [using (i)]
Hence angle made by velocity vector of image is $-\frac{\pi }{4}$