A long narrow slit is illuminated by blue light and the diffraction pattern is obtained on a white screen. What do you mean by limit of resolution of an optical instrument ?

The limit of resolution (or resolving power) is a measure of the ability of the objective lens to separate in the image adjacent details that are present in the object.It is the distance between two points in the object that are just resolved in the image.The resolving power of an optical system is ultimately limited by diffraction by the aperture.Thus an optical system cannot form a perfect image of a point.

The figure shows the image formation of an object in simple microscope. Derive an equation for magnifying power of the simple microscope.

Magnifying power is the ratio of the angle subtended at the eye by the image at the near point, to the angle subtended by the object, at least distance of distinct vision. In the given figure, the image is formed at the near point. According to the figure, the angle subtended by the image at the eye is β and the angle subtended by the object at the unaided eye is α . Let the object distance be u = O Q and the image distance be D = O Q 1 . Hence, the magnifying power can be given as: M = Angle subtended by the image at the eye through the lens Angle subtended by the object at least distance of distinct vision ⇒ M = β α From the triangle P Q O , tan β = P Q Q O ⇒ β = P Q Q O ∵ β is very small . . . 1 From the triangle P Q 1 O , tan α = P Q 1 O Q 1 ⇒ α = P Q 1 O Q 1 ∵ α is very small ⇒ α = P Q O Q 1 ∵PQ 1 =PQ . . . 2 From equations 1 and 2 , the magnifying power can be calculated as: M = β α ⇒ M = P Q Q O × O Q 1 P Q ⇒ M = O Q 1 O Q = D u From the lens formula: 1 f = 1 v - 1 u 1 f = - 1 v + 1 u ( from sign convention ) - 1 D + 1 u = 1 f ⇒ - 1 + D u = D f ⇒ 1 + D f = D u ⇒ M = 1 + D f ∵ M = D u Hence, the formula for magnifying power of a simple microscope is M = 1 + D f .

The figure shows the image formation of an object in simple microscope. Find out the object distance and image distance from the figure.

According to the diagram given in the question, the rays originate from the point P . Hence, the object is placed on the point P . The object distance is the distance of the object to the centre line of the lens. Hence, O Q is the object distance. The rays after refracting from the lens and passing through the centre of the lens, meet at the point P 1 . The image distance is the distance of the image to the centre line of the lens. Hence, O Q 1 is the image distance.