Question

A concave mirror forms a real image 4 times the size of an object. The magnification is 3 times by moving the object 3 cm away from the mirror. Find out the radius of curvature of the mirror.

Hint:

For mirrors, magnification is given by \( m = -\frac{v}{u} \). Use the mirror formula to relate object distance, image distance, and focal length.

Answer 1

For a concave mirror, the magnification \( m = -\frac{v}{u} \), where \( v \) is the image distance and \( u \) is the object distance. Case 1 (\( m = -4 \)): \[ v = -4u. \] The mirror formula is: \[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v}. \] Substituting \( v = -4u \): \[ \frac{1}{f} = \frac{1}{u} + \frac{1}{-4u} = \frac{3}{4u}. \] Case 2 (\( m = -3 \), object moved by 3 cm): \[ u' = u + 3, \quad v' = -3u'. \] Using the mirror formula: \[ \frac{1}{f} = \frac{1}{u'} + \frac{1}{v'}. \] Substituting \( v' = -3u' \): \[ \frac{1}{f} = \frac{1}{u+3} + \frac{1}{-3(u+3)} = \frac{2}{3(u+3)}. \] Equating \( \frac{1}{f} \) from both cases: \[ \frac{3}{4u} = \frac{2}{3(u+3)}. \] Solving for \( u \): \[ u = 12 \, \mathrm{cm}, \quad f = 9 \, \mathrm{cm}. \] The radius of curvature is: \[ R = 2f = 18 \, \mathrm{cm}. \]