KVPY SA Exam 2015 Question Paper

Question : 69 of 80

Marks: +1, -0

A point object is placed 20 cm left of a convex lens of focal length f = 5 cm (see the figure). The lens is made to oscillate with small amplitude A along the horizontal axis. The image of the object will also oscillate along the axis with

amplitude $A / 9$ , out of phase with the oscillations of the lens.
amplitude $A/3$ , out of phase with the oscillations of the lens.
amplitude $A /3$ , in phase with the oscillations of the lens.
amplitude $A/9$ , in phase with the oscillations of the lens

Validate

Solution:

1/\f=1/v-1/u

1/v=1/\f+ 1/u

v={\fu}/{\f+u}

m=v/u=\f/{\f+u}

As lens is oscillating with small amplitude A.
∴ Image will oscillate with m

^{2}

A
When lens move left then O will come near to lens thus I will go away. Thus image is oscillating out of phase with respect to lens.

m=5/{5-20}⇒5/{-15}=-1/3

Amplitude of image =

(1/3)^2\;\;\;A=A/9

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