Question

63. The radius of curvature of the curved face of a thin planoconvex lens is 10 cm and it is made of glass of refractive index 1.5. A small object is approaching the lens with a speed of 1 cms' moving along the principal axis Lens A B F C D Object M2 M3 a When the object is at a distance of 30 cm from the lens, the magnitude of the rate of change of the lateral magnification is (B) (A) 0.1 per second (C) 0.3 per second 0.2 per second 0.4 per second (D)​

Answer 1

Answer:

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Answer 2

Given:

The radius of curvature of the planoconvex lens (f) = 10 cm

The refractive index of the lens (μ) = 1.5

The velocity of the approaching object = 1 cm /s

To Find:

The magnitude of the rate of change of the lateral magnification (dm/dt) when the object is at 30 cm

Solution:

Using Lens Formula,

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

Substituting u = -30 and f = 10,

⇒(1 / v) - (-1/30) = 1 / 10

or v = 15 cm

Image velocity dv / dt = (v/u)² X du/dt

Substituting  v = 15 cm, u = -30cm and du/dt = 1 cm/s (object velocity)

Image velocity = dv/dt = 1/4 cm/s

Magnification formula, m =  v / u

Differentiating wrt 't',

$\frac{dm}{dt} = \frac{u\frac{dv}{dt}-v\frac{du}{dt} }{u^2}$

Substituting the values,

dm/dt = -30X0.25 - 15X1 / 900

= - 0.025 ≈ -0.1 per second

Hence, The magnitude of the rate of change of the lateral magnification is (a) 0.1 per second.