Question

a point object is placed on the axis of a thin convex lens of focal length 0.05m at a distance of 0.02m from the lens and its image is formed on the axis. if the object is now made to oscillate along the axis with a small amplitude of a cm then what is the amplitude of oscillation of image​

Answer 1

Explanation:

Given A point object is placed on the axis of a thin convex lens of focal length 0.05 m at a distance of 0.2 m from the lens and its image is formed on the axis. if the object is now made to oscillate along the axis with a small amplitude of A cm then what is the amplitude of oscillation of image

• Given u = - 0.2 m, f = 0.05 m
• We know that 1/v = 1/u + 1/f
•                        1/v = 1/- 0.2 + 1/0.05
•                         1/v = - 5 + 20
•                        1/v = 15
• Or v = 15 m
• Now dv / v^2 = - du / u^2
• Therefore dv = A. v^2 / u^2
• Now A image = A x (1/15)^2 / (- 0.2)^2
•                       = A . 1/225. 1/0.04
•                       = A. 1 / 225 x 0.04
•                       = A. 1/9
•                           = A cm / 9  
•                          = A/9 x 10^-2 m

Reference link will be

https://brainly.in/question/12369003