Rankers Physics

Refraction by Spherical Surfaces: Practice Problem & Solution

Match the corresponding entries of column-1 with column-2. [where m is the magnification produced by the mirror] \begin{array}{ll} \textbf{Column-I} & \textbf{Column-II} \\[1ex] \text{(A) } m = -2 & \text{(1) Convex mirror} \\ \text{(B) } m = -\frac{1}{2} & \text{(2) Concave mirror} \\ \text{(C) } m = +2 & \text{(3) Real image} \\ \text{(D) } m = +\sqrt{\frac{2}{3}} & \text{(4) Virtual image} \end{array} (2016 - I)
A -> 2 and 3; B -> 2 and 3; C -> 2 and 4; D -> 1 and 4
A -> 1 and 3; B -> 1 and 4; C -> 1 and 2; D -> 3 and 4
A -> 1 and 4; B -> 2 and 3; C -> 2 and 4; D -> 2 and 3
A -> 3 and 4; B -> 2 and 4; C -> 2 and 3; D -> 1 and 4

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Refraction by Spherical Surfaces. Understanding the underlying formula is key to arriving at the correct answer below:

Negative magnification implies a real inverted image, which is only formed by a concave mirror. So A ( $m = -2$ ) and B ( $m = -1/2$ ) match with (2) Concave and (3) Real. Positive magnification > 1 ( $m = +2$ ) implies a virtual magnified image, formed only by a concave mirror (2, 4). Positive magnification < 1 implies a virtual diminished image, formed by a convex mirror (1, 4).

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