Reflection Of Light: Spherical Mirrors

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Question
ICSEENIPH12030010

Using Huygen’s Principle, prove the laws of refraction of light. 

Solution

Huygen’s principle provides a geometrical method to determine the position of the wavefront at a later time from its given position at any instant.
It states that:

(1) Each point on a wavefront behaves as a new source of secondary wavelets, which spreads out in all the directions with the speed of light in the medium .

(2) The new position of the wavefront at any instant is given by the surface of tangents to the secondary wavelets in the forward direction. To explain the phenomenon of refraction by Huygen’s principle, let us consider the surface XY separating the media 1 and 2 of refractive indices μ2 and μ2, respectively and c1 and c2 be the velocities of light in the two media.
The second medium being more denser than the first and so c1 will be greater than c2 i.e., c1 > c2.
APB is the plane incident wave front. Let t be the time taken by the waves to travel the distance BC, the BC = c1t and AD = c2t. With A as centre and radius AD, draw a sphere and a tangent CD to the sphere from the point C. Then represents the refracted plane wave front.
CD is the common wave front since in the time the disturbance travels from B to C or from A to D, the disturbance at P reaches L. With point M as centre, draw a sphere such that CD happens to be tangent to the sphere.

From the As ACD and MCL, we have
AD over ML space equals space fraction numerator A C over denominator M C 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 left parenthesis 1 right parenthesis

S i m i l a r l y comma space I n space increment space A C E space a n d space M C N comma

fraction numerator A E over denominator M N end fraction space equals space fraction numerator A C over denominator M N 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 left parenthesis 2 right parenthesis

F r o m space left parenthesis 1 right parenthesis space a n d space left parenthesis 2 right parenthesis comma space w e space h a v e

space space space space space space space fraction numerator space space space A D over denominator M L end fraction space equals space fraction numerator A E over denominator M N end fraction

rightwards double arrow space space space space space fraction numerator A E over denominator A D end fraction space equals space fraction numerator B C over denominator A D end fraction space equals space fraction numerator c subscript 1 t over denominator c subscript 2 t end fraction space equals space fraction numerator M N over denominator M L end fraction

space space space space space space space space fraction numerator B C over denominator A D end fraction space equals space fraction numerator M N over denominator M L end fraction space equals space c subscript 1 over c subscript 2
Hence, AD is the radius for secondary wave front for the point A, then ML is the radius of the secondary wave front for the point M. Let i and r be the angle of incidence and refraction, respectively. From the Δs ABC and ACD.
fraction numerator sin space straight i over denominator sin space straight r end fraction space equals space fraction numerator begin display style bevelled fraction numerator B C over denominator A C end fraction end style over denominator begin display style bevelled fraction numerator A D over denominator A C end fraction end style end fraction space equals space fraction numerator B C over denominator A D end fraction space equals space fraction numerator c subscript 1 t over denominator c subscript 2 t end fraction space equals space c subscript 1 over c subscript 2 space equals space mu subscript 1 over mu subscript 2
where μ2 is the refractive-index of the second medium with respect to the first medium. The above relation is Snell’s law.

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Question
ICSEENIPH12030273

A point object O is placed at a distance of 15 cm from a convex lens L of focal length 10 cm as shown in figure below. On the other side of the lens, a convex mirror M is placed such that its distance from the lens is equal to the focal length of the lens. The final image formed by this combination is observed to coincide with the object O. Find the focal length of the convex mirror.

Solution

Given,
Object distance, u = 15 cm
Focal length, f = 10 cm
Using the lens formula, we have
space space space space space space 1 over 15 plus 1 over v equals 1 over 10

therefore space v space equals space 30 space c m

space space space 2 f space equals space R space equals space v minus 10

space space space 2 f space equals space 30 space minus space 10
rightwards double arrow space f space equals space 10