Optics 7 Question 46

47. The convex surface of a thin concavo-convex lens of glass of refractive index 1.5 has a radius of curvature 20cm. The concave surface has a radius of curvature 60cm. The convex side is silvered and placed on a horizontal surface.

(1981, 2M)

(a) Where should a pin be placed on the optic axis such that its image is formed at the same place?

(b) If the concave part is filled with water of refractive index 4/3, find the distance through which the pin should be moved, so that the image of the pin again coincides with the pin.

Show Answer

Solution:

  1. (a) Image of object will coincide with it if ray of light after refraction from the concave surface fall normally on concave mirror so formed by silvering the convex surface. Or image after refraction from concave surface should form at centre of curvature of concave mirror or at a distance of 20cm on same side of the combination. Let x be the distance of pin from the given optical system.

Applying, μ2vμ1u=μ2μ1R

With proper signs, 1.5201x=1.5160

or 1x=3401120=8120x=1208=15cm

(b) Now, before striking with the concave surface, the ray is first refracted from a plane surface. So, let x be the distance of pin, then the plane surface will form its image at a distance 43x(happ. =μh) from it.

Now, using μ2vμ1u=μ2μ1R with proper signs,

we have 1.5204/34x3=1.54/360

or 1x=3401360 or x=13.84cm

Δx=x1x2=15cm13.84cm

=1.16cm

(downwards)

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