## Monday, July 07, 2008

### Multiple Choice Questions involving Lens Maker’s Equation

While attempting to solve problems in optics related to spherical mirrors and lenses, many among you will have some confusion about the application of the inevitable sign convention. Get your confusion cleared by going through the post on this site here. You can find all posts in optics on this site by clicking on the label ‘OPTICS’ below this post. Your doubts regarding the application of the sign convention can be cleared completely only by working out questions involving the sign convention.

We will now discuss a few questions involving lens maker’s equation:

(1) Crown glass of refractive index 1.5 is used to make a plano concave lens of focal length 40 cm in air. What should be the radius of curvature of the curved face?

(a) 20 cm

(b) 40 cm

(c) 60 cm

(d) 80 cm

(e) 120 cm

This is a situation where you have to use the lens maker’s equation,

1/f = (n2/n1 – 1)(1/R1 – 1/R2) where ‘f’ is the focal length of the lens, R1 and R2 are its radii of curvature, n2 is its refractive index and n1 is the refractive index of the medium in which the lens is placed.

Since the lens is plano concave and hence diverging, its focal length is negative according to the Cartesian sign convention. The above equation on substituting the known values (n1 = 1, n2 = 1.5 and R2 = ∞) becomes

–1/40 = (1.5 – 1)( 1/R1 – 0)

This gives R1 = –20 cm [Option (a)].

You obtain the answer as negative in accordance with the sign convention, indicating that the curved face is concave towards the incident ray. If you had taken the second face as curved, you would have put R1 = ∞ and the answer would be R2 = 20 cm. This too conforms to the Cartesian sign convention, indicating that the second face is convex towards the incident ray and hence the radius of curvature is positive.

In the present problem, there will be no need of worrying too much about the sign convention if you remember that the focal length (in air) and the radius of curvature will have the same value in the case of biconvex and biconcave lenses of equal radii of curvature if the refractive index of the lens is 1.5. In the case of plano convex and plano concave lenses made of material of refractive index 1.5, the radius of curvature of the curved face will be half the value of the focal length.

(2) A biconvex lens has the same radii of curvature for its faces. If its focal length in air is equal to the radii of curvature of its faces, its focal power when immersed in a liquid of refractive index 1.5 will be (in dioptre)

(a) 0.667
(b) 1.5
(c) 0.5
(d) zero
(e) data insufficient.
You may substitute the same values for the focal length in air and the radii of curvature of the faces in the lens maker’s equation and satisfy yourself that the refractive index of the lens is 1.5.[Since it is a biconvex lens, f is positive, R1 is positive and R2 is negative. Also, put the numerical values of R1 and R2 equal to f].Since the refractive index of the liquid is the same as that of the lens, there is no convergence or divergence for the rays of light and the focal power will be zero.
(3) The radius of curvature of the convex face of a plano convex lens is 15 cm and the refractive index of the material is 1.4. Then the power of the lens in dioptre is
(a) 1.6
(b) 1.566
(c) 2.6
(d) 2.66
(e) 1.4
The above question appeared in Kerala Engineering Entrance 2008 question paper.

You have to use the lens maker’s equation,

1/f = (n2/n1 – 1)(1/R1 – 1/R2) where ‘f’ is the focal length of the lens, R1 and R2 are its radii of curvature, n2 is its refractive index and n1 is the refractive index of the medium in which the lens is placed.

Note that you have to substitute the focal length in metre in this equation which gives you the focal power 1/f itself. Substituting the given values,

1/f = (1.4 – 1)(1/0.15 – 0) = 2.66 nearly.

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