Prove that for a prism μ = sin A + δ m 2 sin A 2 Where symbols have their usual.

Let the refractive index of the prism be μ Applying Snell's law, n air sin i = μ sinr where n air = 1 Or sini = μ sin r 1 ⇒ μ = sin i sin r 1 . . . . . . . . 1 Deviation by prism δ = i + e − A where A = r 1 + r 2 For minimum deviation i = e and r 1 = r 2 ∴ A = 2 r 1 ⇒ r 1 = A 2 . . . . . . b Angle of minimum deviation is given by δ m = 2 i − A ⇒ i = A + δ m 2 . . . . . c Putting ( b ) and ( c ) in ( 1 ) , we get μ = sin A + δ m 2 sin A 2

Deduce Lens Makers formula for a double convex lens 1 f = μ - 1 1 R 1 - 1 R 2 Where the symbols have their usual meanings.

Let a biconvex lens is made up of two surfaces AB and CD of radii of curvatures R 1 and R 2 . A point object is placed on principal axis at O. Image of an object O is formed at I due to AB and this image acts as an object for second surface CD and its image produced at I 1 Let distance of object O from the lens is u image I 1 is at distance v 1 , image I is at distance v From first surface AB image of O get on I 1 that way by refraction formula μ - 1 R 1 = μ v 1 - 1 u . . . . . . ( i ) From second surface CD image of I 1 get on I by the refraction formula 1 μ - 1 R 2 = 1 μ - 1 v - 1 v 1 . . . . . . i i μ = 1 u , u = v 1 for CD surface because light ray passing from denser to rarer medium Equation 2 multiply by μ 1 v - μ v 1 = 1 - μ R 2 . . . . iii Add equation 1 + 3 1 v - 1 μ = μ - 1 R 1 + 1 - μ R 2 . . . . iv 1 v - 1 u = μ - 1 R 1 - μ - 1 R 2 if = ∞ , v = f 1 f - 1 ∞ = μ - 1 1 R 1 - 1 R 2 1 ∞ = 0 1 f = μ - 1 1 R 1 - 1 R 2

The radii of curvature of the faces of a double convex lens made of glass, are 10 cm and 15 cm . Its focal length is 12 cm . a Find the refractive index of glass b Calculate its new focal length if the lens is completely immersed in water ( Refractive index of water is 4 3 )

1 f = μ - 1 1 r 1 - 1 r 2 1 12 = μ - 1 1 - 10 - 1 15 Therefore μ = - 0 . 5

Draw a neat ray diagram showing the formation of final image by a compound microscope.

Compound microscope is consist of two convex lenses, one object of very small focal length with short aperture and one eyepiece, E of moderate focal length and large aperture. The ray diagram showing the formation of image by a compound microscope is shown in the diagram.

A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)

Given data, Actual Depth of bulb in water tank, d = 80 cm = 0 . 8 m The refractive index of the water, μ w = 1 . 33 Since, the bulb is considered as point source, therefore the light coming out from surface of water will be in circle also known as the circle of illuminance. The light ray from the source incident with a critical angle on the border of this circle. The diameter of the circle will be AC (from the above image), and circle radius = R , From image, sin θ i = R R 2 + 80 2 From Snell’s law, μ w sin ( θ i ) = μ a sin 90 ⇒ 1.33 × R R 2 + 80 2 = 1 ⇒ R = 91 cm = 0 . 91 m Area π R 2 = 2.61 m 2

Deduce Lens Makers formula for a double convex lens 1 f = μ - 1 1 R 1 - 1 R 2 Where the symbols have their usual meanings.

Let a biconvex lens is made up of two surfaces AB and CD of radii of curvatures R 1 and R 2 . A point object is placed on principal axis at O. Image of an object O is formed at I due to AB and this image acts as an object for second surface CD and its image produced at I 1 Let distance of object O from the lens is u image I 1 is at distance v 1 , image I is at distance v From first surface AB image of O get on I 1 that way by refraction formula μ - 1 R 1 = μ v 1 - 1 u . . . . . . ( i ) From second surface CD image of I 1 get on I by the refraction formula 1 μ - 1 R 2 = 1 μ - 1 v - 1 v 1 . . . . . . i i μ = 1 u , u = v 1 for CD surface because light ray passing from denser to rarer medium Equation 2 multiply by μ 1 v - μ v 1 = 1 - μ R 2 . . . . iii Add equation 1 + 3 1 v - 1 μ = μ - 1 R 1 + 1 - μ R 2 . . . . iv 1 v - 1 u = μ - 1 R 1 - μ - 1 R 2 if = ∞ , v = f 1 f - 1 ∞ = μ - 1 1 R 1 - 1 R 2 1 ∞ = 0 1 f = μ - 1 1 R 1 - 1 R 2