Triangle on a concave mirror

JEE Advanced 2018 Paper 2, Question 4

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.)

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Related article: Optics without ray diagrams: Mirrors

Solution

If the object is kept between a concave mirror and its focal point the image is virtual and upright (see this note). Recall that an object at a distance $p$ from a mirror of focal length $f$ forms an image at $q$ , such that

(1) $\begin{equation*} \frac{1}{p} + \frac{1}{q} = \frac{1}{f} . \end{equation*}$

The given object is extensive, that is, different points of the object are located at different $p$ . The point $B$ is at $p=f$ (see figure), which means there will be a portion of the image at $q = -\infty$ . Therefore, the correct option must be either (B) or (D). To narrow it down further let us consider the segment $AB$ . The point $A$ is $(h_0,f/2)$ and $B$ is $(0,f)$ (the triangle is isoceles, which means $h_0=f/2$ , but we will continue calling this height $h_0$ for the time being). An arbitrary point on the segment $AB$ is given by

(2) $\begin{equation*} \frac{h-y_B}{p-x_B} = \frac{y_A-y_B}{x_A-x_B} \implies h = \frac{2 h_0}{f} (f-p) . \end{equation*}$

This point will be located at a distance $q$ given by (1),

(3) $\begin{equation*} q = \frac{f p}{p-f} , \end{equation*}$

with a height $h' = M \times h$ , where $M = -q/p$ is the magnification. That is,

(4) $\begin{equation*} h' = \frac{-\frac{f p}{p-f}}{p} \times \frac{2 h_0}{f} (f-p) = 2 h_0 , \end{equation*}$

which is a constant. Therefore all points on $AB$ appear at the same height in the image. Therefore, the correct answer is (D).

Discussion: The points on the pricipal axis between $f/2$ and $f$ will also be be in the image, although this is not clear from the picture given in the exam. A better illustration would be the figure below.

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Also note that $h' = 2 h_0$ for any height $h_0$ (within the field of view of the mirror, of course). That is, the image has the shape shown above for any $\angle B$ , and not just $\angle B = 45^\circ$ .

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