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 .

A small angled prism of 4 ° deviates a ray through 2 . 48 ° . Find the refractive Index of the prism.

It is given that the prism under consideration is a thin prism. Also, The angle of prism, A = 4 ° Angle of deviation, δ = 2 . 48 ° Let n be the refractive index of the prism. We know that for a thin prism, δ = A n - 1 ⇒ n = δ A + 1 ⇒ n = 1 . 62

A concave mirror produces an image of a long vertical pin, placed 40 cm from the mirror, at the position of the object. Find the focal length of the mirror (in cm ).

When an object is placed at the centre of curvature of a concave mirror, a real, inverted image of the same size as the object will form at the centre of curvature itself. This is illustrated in the figure. It is given in the question that image is formed at the same position as the object. This implies that the object is placed at the centre of curvature. That means, radius of curvature, R = - 40 cm Therefore, focal length, f = - 40 2 ⇒ f = - 20 cm