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Wave Optics

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(i) Derive the mathematical relation between refractive indices n₁ and n₂ of two radii and radius of curvature R for refraction at a convex spherical surface. Consider the object to be a point since lying on the principle axis in rarer medium of refractive index n₁ and a real image formed in the denser medium of refractive index n₂. Hence, derive lens maker’s formula.

(ii) Light from a point source in air falls on a convex spherical glass surface of refractive index 1.5 and radius of curvature 20 cm. The distance of light source from the glass surface is 100 cm. At what position is the image formed ?

Solution

i)

The figure given above shows the geometry of formation of real image I of an object O and the principal axis of a spherical surface with centre of curvature c and radius of curvature R.

Assumptions:
1. The aperture of the surface is small compared to other distance involved.
2. NM is taken to be nearly equal to to the length of the perpendicular from the point N on the principal axis.

tan space angle N O M space equals space fraction numerator M N over denominator O M end fraction tan space angle N C M space equals space fraction numerator M N over denominator M C end fraction tan space angle N I M space equals space fraction numerator M N over denominator M I end fraction F o r space increment N O C comma i space i s space t h e space e x t e r i o r space a n g l e.

Assuming the incident ray is very close to the principal axis, all the angles are very small. Hence, for small angles,
tan x = x = sin x
Therefore,

straight i space equals space angle NOM plus angle NCM straight i space equals space MN over OM plus MN over MC space bold space bold. bold. bold. bold space bold left parenthesis bold i bold right parenthesis Similarly comma space straight r space equals space angle NCM space minus space angle NIM straight r equals space MN over MC space minus space MN over MI space bold space bold. bold. bold. bold space bold left parenthesis bold ii bold right parenthesis Using space Snell apostrophe straight s space law comma straight n subscript 1 space sin space straight i space equals space straight n subscript 2 space sin space straight r For space small space angles comma straight n subscript 1 straight i space equals space straight n subscript 2 space straight r

Putting the values of i and r from eqns. from (i) and (ii), we have

straight n subscript 1 open parentheses MN over OM plus MN over MC close parentheses space equals space straight n subscript 2 open parentheses MN over MC minus MN over MI close parentheses straight n subscript 1 over OM plus straight n subscript 2 over MI equals fraction numerator space straight n subscript 2 minus straight n subscript 1 over denominator MC end fraction space bold. bold. bold. bold space bold left parenthesis bold iii bold right parenthesis

Applying new cartesian sign conventions,
OM = -u, MI = + v, MC = +R
Putting these values in equation (iii), we have

straight n subscript 2 over straight v minus n subscript 1 over u space equals space fraction numerator n subscript 2 minus n subscript 1 over denominator R end fraction

ii) According to the question,

space space space straight n subscript 2 over straight v minus n subscript 1 over u equals fraction numerator n subscript 2 minus n subscript 1 over denominator R end fraction rightwards double arrow space fraction numerator 1.5 over denominator V end fraction minus fraction numerator 1 over denominator left parenthesis negative 100 right parenthesis end fraction equals fraction numerator 1.5 minus 1 over denominator 20 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1 over 40 rightwards double arrow space fraction numerator 1.5 over denominator V end fraction equals 1 over 40 minus 1 over 100 space space space space space space space space space space space space space space equals fraction numerator 5 minus 2 over denominator 200 end fraction space space space space space space space space space space space space space space equals space 3 over 200 rightwards double arrow space v space equals space fraction numerator 1.5 space x 200 over denominator 3 end fraction space space space space space space space space space equals 100 space c m

Therefore, the image is formed at a distance of 100 cm in the denser medium.

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