Question

A point object at 15 cm from a concave mirror of radius of curvature 20 cm is made to oscillate along the principal axis with amplitude 2 mm. The path length of the image in mm is
(A) 2mm
(B) 4mm
(C) 8mm
(D) None of these

Answer 1

Hint : We need to apply Lens makers’ formula. Using the relationship between focal length, image and object distance we will be able to calculate the image distance from the formula, $ \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} $ .

Formula Used: The following formulas are used to solve this question.
 $ \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} $ where $ f $ is the focal length, $ u $ is the object distance, $ v $ is the image distance.
Transverse magnification $ {M_T} = - \dfrac{v}{u} $
Longitudinal magnification $ {M_L} = - \dfrac{{dv}}{{du}} = {M_T}^2 $

Complete step by step answer
Spherical mirrors can be thought of as a portion of a sphere that was sliced away and then silvered on one of the sides to form a reflecting surface. Concave mirrors are silvered on the inside of the sphere. The point on the mirror's surface where the principal axis meets the mirror is known as the vertex and the distance from the vertex to the center of curvature is known as the radius of curvature. The radius of curvature is the radius of the sphere from which the mirror was cut. Since the focal point is the midpoint of the line segment adjoining the vertex and the center of curvature, the focal length would be one-half the radius of curvature.
Since the radius of curvature is 20cm, focal length is 10cm.
Given that, object distance $ u = 15cm $ and focal length $ f = 10cm $ .
The object oscillates along the principal axis with amplitude 2 mm. Thus, $ du = 2mm. $
According to Lens Makers’ Formula,
 $ \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} $ where $ f $ is the focal length, $ u $ is the object distance, $ v $ is the image distance.
Assigning, $ u = 15cm $ and focal length $ f = 10cm $ ,
 $ \dfrac{1}{v} + \dfrac{1}{{ - 15}} = \dfrac{1}{{ - 10}} $
 $ \Rightarrow v = - 30cm. $
Thus the image distance is -30cm.
Transverse magnification $ {M_T} = - \dfrac{v}{u} $
Assigning the values,
 $ {M_T} = - \dfrac{{\left( { - 30} \right)}}{{\left( { - 15} \right)}} = - 2 $
Longitudinal magnification $ {M_L} = - \dfrac{{dv}}{{du}} = {M_T}^2 $
Transverse magnification is given by $ - 2 $ . Thus, longitudinal magnification, $ {M_L} = {M_T}^2 = - {\left( { - 2} \right)^2} $
 $ \Rightarrow {M_L} = - 4 $
Now since, $ {M_L} = - \dfrac{{dv}}{{du}} = - 4 $
 $ \Rightarrow - \dfrac{{dv}}{{du}} = - 4 $
Here we assign the value $ du = 2mm. $
 $ \Rightarrow dv = 8mm. $
Thus, the image will oscillate with an amplitude of 8mm.
The correct answer is Option C.

Note
Magnification is the process of enlarging the apparent size, not physical size, of something. This enlargement is quantified by a calculated number also called "magnification". For real images, such as images projected on a screen, size means a linear dimension and is called a linear or transverse magnification.