QJEEAdv18Ph046d: JEE Advanced 2018 Physics - Class 12 JEEAdv18Ph046d Solution: A wire is bent in the shape of a right angled triangle and is…

Consider the point F, the focus of the concave mirror. All rays emanating from F will be reflected back parallel to the principal axis PF. Thus, virtual image F' of the point F is formed at $\infty$ .

Consider the point L at an object distance u = $-$ | f | / 2. For convenience, we use symbol $\widehat f$ for | f | i.e., $\widehat f$ = | f |. Use mirror formula, 1/v + 1/u = 1/f, with f = $-$ $\widehat f$ to get the image distance v = $\widehat f$ . Thus, the virtual image L' of the point L is formed on the left side of the mirror at a distance $\widehat f$ .

The image of a point on LF will lie on the line L'F'. Now, consider the point M. Its object distance is u = $-$ $\widehat f$ /2. Thus, the virtual image M' of the point M is also formed at a distance $\widehat f$ from the mirror. The distance of M' above the principal axis is given by h_M' = $-$ (v/u) (LM) = $-$ ( $-$ 2) ( $\widehat f$ /2) = $\widehat f$ (magnification by a spherical mirror).

Now consider a point N on the line segment FM. Let N₁ be the projection of N on the principal axis PF such that FN₁ = x. The object distance is u = $-$ ( $\widehat f$ $-$ x). The mirror formula gives the image distance v = $\widehat f$ ( $\widehat f$ $-$ x) / x. The distance of the image N' above the principal axis is given by h_N' = $-$ (v/u) (N₁N) = ( $\widehat f$ /x) (x) = $\widehat f$ (it is independent of x). Since, N is an arbitrary point on FM, the image of all points on FM will be formed at a distance $\widehat f$ above the optic axis.