(a) A point-object is placed on the principal axis of a convex spherical surface of radius of curvature R, which separates the two media of refractive indices n1 and n2 (n2 > n1). Draw the ray diagram and deduce the relation between the distance of the object (u), distance of the image (v) and the radius of curvature (R) for refraction to take place at the convex spherical surface from rarer to denser medium.
(b) Use the above relation to obtain the condition on the position of the object and the radius of curvature in terms of n1 and n2 when the real image is formed.
OR
(a) Draw a labelled ray diagram showing the formation of image by a compound microscope in normal adjustment. Derive the expression for its magnifying power.
(b) How does the resolving power of a microscope change when
(i) the diameter of the objective lens is decreased,
(ii) the wavelength of the incident light is increased? Justify your answer in each case.
Consider the following diagram. We assume that the aperture of the surface is very small as compared to the other distances involved.

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In ∆NOC, ![]()
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Similarly, ![]()
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Assuming i and r to be very small angles, so that sin(i)≈i and sin(r)≈r
Using Snell’s law, n1i=n2r
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From the diagram, OM = -u, MI = +v and MC = +R
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OR
(a)

The angular magnification of objective lens, m0 = tanβ = h'/h = L/f0
The angular magnification of eyepiece, me = 1 + D/fe
If the final image is formed at infinity, then angular magnification me = D/fe
Thus, total magnification of the compound microscope
M = m0 × me
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(b) Resolving power of a microscope ![]()
On decreasing the diameter, sinθ decreases, hence resolving power decreases.
On decreasing focal length of objective, sinθ will increase. Hence, resolving power increases.
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