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 .

A ray of light parallel to the principal axis of a spherical mirror falls at a point M as shown in the figure below: By drawing a suitable ray diagram, obtain the mirror equation.

Let points P , F , C be pole, focus and centre of curvature of a concave mirror. Object AB is placed on the principal axis. A ray from AB incident at E and then reflected through F . Another ray of light from B incident at pole P and then reflected. These two rays meet at M . The ray of light from point B is passed through C . Draw EN perpendicular to the principal axis. Δ IMF and Δ ENF are similar. i.e, IM - NE = IF NF . . . . . 1 But IF = PI – PF and NF = PF (since aperture is small) Hence, equation 1 can be written as IM NE = PI - PF PF IM AB = PI - PF PF . . . . . 2 ∵ NE = AB ∆ ABP and ∆ IMP are similar IM AB = PI PA . . . . . 3 From equation 2 and equation 3 , we get PI - PF PF = PI PA Applying sign convention, we get PI = - v PF = - F PA = - u Substituting these in equation 4 , we get - v - f - f = - v - u ⇒ v - f vf = 1 u 1 f - 1 v = 1 u ⇒ 1 f = 1 u + 1 v This is called mirror formula or mirror equation. Hence, it is the required answer.

A ray of light parallel to the principal axis of a spherical mirror falls at a point M as shown in the figure below: Identify the type of mirror used in the diagram.

As we know if a hollow sphere is cut into two parts and the outer surface of the hollow sphere is painted, then it becomes a mirror with its inner surface as the reflecting surface. This type of mirror is known as a concave mirror. And If the inner part of the hollow sphere is painted, then its outer surface becomes the reflecting surface. This kind of mirror is known as a convex mirror. So, from the figure we can say that the mirror is concave mirror.

A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5 cm . What is the magnifying power of the telescope for viewing distant objects when the final image is formed at the least distance of distinct vision.

Given that, The Focal length of the objective lens, f 0 = 140 cm and the Focal length of the eyepiece, f e = 5 cm . Also the Least distance of distinct vision, d = 25 cm . Therefore, When the final image is formed at d , its magnifying power is given as: m = f 0 f e 1 + f e d ⇒ m = 140 5 1 + 5 25 = 28 × 1 . 2 = 33 . 6 Hence, 33 . 6 is the required answer.

A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5 cm . What is the magnifying power of the telescope for viewing distant objects when the telescope is in normal adjustment.

Given that, The Focal length of the objective lens, f 0 = 140 cm , and the Focal length of the eyepiece, f e = 5 cm . Also, the Least distance of distinct vision, d = 25 cm . When the telescope is in normal adjustment, its magnifying power is given as m = f 0 f e . So, m = f 0 f e = 140 cm 5 cm = 28 . Hence, 28 , is the required answer.

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 .