Optics Ques 13
- A wire is bent in the shape of a right angled triangle and is placed in front of a concave mirror of focal length $f$ as shown in the figure. Which of the figures shown in the four options qualitatively represent(s) the shape of the image of the bent wire? (These figures are not to scale.)
(2018 Adv.)
(a)
(b)
(c)
(d)
Show Answer
Answer:
Correct Answer: 13.(d)
Solution:
Formula:
Lateral Magnification (Or Transverse Magnification)
- Image of point $A$
$ \frac{P Q}{x}=\frac{A B}{f / 2} \Rightarrow P Q=\frac{2(A B) x}{f} $
For $A$ :
$ \frac{1}{v}+\frac{1}{[-(f / 2)]}=\frac{1}{-f} \Rightarrow v=f $
$ \begin{array}{cc} \Rightarrow & \frac{I _{A B}}{A B}=-\frac{v}{u}=-\frac{f}{(-\frac{f}{2})} \\ \Rightarrow & I _{A B}=2 A B \end{array} $
For height of $P Q$,
$ \begin{aligned} & \frac{1}{v}+\frac{1}{-[(f-x)]}=\frac{1}{-f} \\ & \Rightarrow \quad \frac{1}{v}=\frac{1}{(f-x)}-\frac{1}{f} \Rightarrow \quad v=\frac{f(f-x)}{x} \\ & \Rightarrow \quad \frac{I _{P Q}}{P Q}=-\frac{v}{u}=\frac{f(f-x)}{x[(f-x)]}=(\frac{f}{x}) \\ & \Rightarrow \quad I _{P Q}=\frac{f}{x} P Q=(\frac{f}{x} ) \quad (\frac{2(A B) x}{f}) \quad [\because P Q=\frac{2(A B) x}{f}] \\ & I _{P Q}=2 A B \end{aligned} $
(Size of image is independent of $x$. So, final image will be of same height terminating at infinity.)
BETA
