The radii of curvature of the surfaces of a thin lens are +10cm and +30 cm. The index is 1.50.
(a) Compute the position and size of the image of an object in the form of an arrow 1 cm high, perpendicular to the lens axis, 40 cm to the left of the lens.
(b) A second similar lens is placed 160 cm to the right of the first. Find the position of the final image.
(c) Same as (b) except the second lens is 40 cm to the right of the first.
(d) Same as (c) except the second lens is diverging, of focal length -40cm.

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Solution: We use the sign convention: i) All distances are measured from the pole of the lens and ii) all distances measured in the direction of the incident ray are positive and those directed opposite are negative.

We note that the radii of curvature of both the surfaces of the lens are positive. As per the sign convention: i) positive radius of curvature of the first surface implies it is convex and ii) positive radius of curvature of the second surface implies it is concave. Hence, the lens is a convexo-concave one.

First we compute the focal length of the lens using the lens maker's formula.
Lens maker's formula: 1/f = (n - 1)[1/R1 - 1/R2]
Substituting n = 1.5, R1 = +10 cm and R2 = +30 cm we get:
1/f = (1.5 - 1)[1/10 - 1/30] = 0.0333
=> f = +30 cm

a) We can compute the required values using the lens formula viz. 1/v - ...

Solution Summary

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