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Question

An object lying at the distance of 5 cm from a spherical interface as shown in the figure, is moving with the velocity of 5 cm/s. If the interface is moving with a velocity of 3 cm/s, then calculate the velocity of the image formed by the spherical surface.
(Assume that spherical interface always seperates air and medium of refractive index 1.5)


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Solution

Given :
Velocity of object (VO)=5 cm/s
Velocity of spherical interface (VS)=3 cm/s
Distance of object from the pole of the interface (u)=5 cm
Radius of the spherical interface (R)=5 cm
Referactive index of the medium (μ2)=1.5

Let the velocity of the image formed is VI

We know that for spherical interface, position of the image is given by :

μ2vμ1u=μ2μ1R

1.5v15=1.515

v=15 cm

We know that the velocity of the image is given by :

VIS=μ1μ2.v2u2.VOS

VIVS=μ1μ2.v2u2.(VOVS)

VI(3)=11.5×(15)2(5)2×[5(3)]

VI+3=2251.5×25×8

VI+3=48

VI=45 cm/s


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