With a neat labelled diagram explain the formation of image in a simple microscope.

A simple magnifier or microscope is a converging lens of small focal length. If the object is at a distance slightly less than the focal length of the lens, the image is virtual and closer than infinity. The near point of the normal eye is 25 cm . i.e. it causes some strain on the eye. Therefore, the image formed at infinity is often considered most suitable for viewing by the relaxed eye. The linear magnification m for the image formed at the near point D by a simple microscope can be obtained by using the relation m = v u = v 1 v - 1 f = 1 - v f Now according to our sign convention, v is negative, and is equal in magnitude to D Thus, the magnification is m = 1 + D f We will now find the magnification when the image is at infinity. In this case we will have to obtain the angular magnification. Suppose the object has a height h . The maximum angle it can subtend, and be clearly visible (without a lens), is when it is at the near point, i.e. at a distance D . The angle subtended is then given by tan θ o = h D ≈ θ o The angle subtended at the eye by the image when the object is at u . From the relations h ' h = m = v u We have the angle subtended by the image, tan θ i = h ' - v = h - v v u = h - u ≈ θ i ⇒ θ i = h f (when u = - f ) Therefore, angular magnification is m = θ i θ o = D f .

Define focal length of a concave mirror. Prove that the radius of curvature of a concave mirror is double its focal length.

The focal length of a concave mirror is the distance between the Pole and the principal focus. Consider the following figure of a concave mirror. Let C be the centre of curvature of the mirror. Consider a ray parallel to the principal axis striking the mirror at M . Then CM will be perpendicular to the mirror at M . Let θ be the angle of incidence, and MD be the perpendicular from M on the principal axis. Then, ∠ M C P = θ Also, ∠ M F P = 2 θ Now, tan θ = MD CD and tan 2 θ = MD FD For small θ which is true for paraxial rays, tan θ = θ and tan 2 θ = 2 θ Therefore, MD FD = 2 MD CD ⇒ FD = DC 2 Now, for small θ , the point D is very close to the point P . Therefore, FD = f and CD = R That means, f = R 2