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Wave Optics

Question
CBSEENPH12039123

(a) Obtain lens makers formula using the expression

 n subscript 2 over v minus n subscript 1 over u equals fraction numerator left parenthesis n subscript 2 minus n subscript 1 right parenthesis over denominator R end fraction

Here the ray of light propagating from a rarer medium of refractive index (n1) to a denser medium of refractive index (n2) is incident on the convex side of spherical refracting surface of radius of curvature R.

(b) Draw a ray diagram to show the image formation by a concave mirror when the object is kept between its focus and the pole. Using this diagram, derive the magnification formula for the image formed.

Solution

Len’s Makers formula:

Consider a thick lens L. The refractive index of the material of lens is n2. The lens is placed in a medium of refractive index n1. R1 and R2 is the radius of curvature of the lens and poles are represented by P1 and P2. Thickness of the lens is given by t. O is a point object which is placed on the principal axis of the lens.

Object distance from pole P1 is u.

Image formed is at a distance of v’ from P1.
Now, using the refraction formula at spherical surface, we have 
fraction numerator straight n subscript 2 over denominator straight v straight apostrophe end fraction minus straight n subscript 1 over straight u equals fraction numerator straight n subscript 2 straight space minus straight space straight n subscript 1 over denominator straight R end fraction                                                 ... (1)

Image I’ acts as a virtual object for second surface and after refraction at second surface, final image is formed at I.

Distance of I from pole P2 of second surface is v.

Distance of virtual object I’ from pole P2 is (v’ – t).
For refraction at second surface, the ray is going from second medium (refractive index n2) to first medium (refractive index n1), therefore from refraction formula at spherical surface.
i.e.,  n subscript 1 over v minus fraction numerator n subscript 2 over denominator left parenthesis v apostrophe minus t right parenthesis end fraction equals fraction numerator n subscript 1 space minus space n subscript 2 over denominator R subscript 2 end fraction

Thickness of the lens is negligibly small as compared to v’. 
Therefore, 
n subscript 1 over v minus fraction numerator n subscript 2 over denominator left parenthesis v apostrophe right parenthesis end fraction equals negative fraction numerator n subscript 2 space minus space n subscript 1 over denominator R subscript 2 end fraction                              ... (2)

Now, adding equations (1) and (2), we have 
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Now, using the lens formula, we get 
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Error converting from MathML to accessible text. is the required Lens maker’s formula. 

b) Ray diagram for the image formation of an object between focus (F) and pole (P) of a concave mirror is given by,

Magnification, m =fraction numerator Size straight space of straight space image left parenthesis straight A straight apostrophe straight B straight apostrophe right parenthesis over denominator Size straight space of straight space object straight space left parenthesis AB right parenthesis end fraction 
From fig. angle space A P B space equals space angle B P Q space equals space i
Also, angle space A P B space equals space angle B P Q space equals space i
I n space increment A P B comma space tan space i space equals space fraction numerator A B over denominator B P end fraction italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic. italic. italic. italic space italic left parenthesis italic 1 italic right parenthesis
I n space increment A apostrophe P B apostrophe comma space tan space i space equals fraction numerator A apostrophe B apostrophe over denominator B apostrophe P end fraction 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 thin space
N o w italic comma italic space f r o m italic space e q u a t i o n italic space italic left parenthesis italic 1 italic right parenthesis italic space a n d italic space italic left parenthesis italic 2 italic right parenthesis italic comma italic space w e italic space h a v e

fraction numerator A B over denominator B P end fraction equals fraction numerator A apostrophe B apostrophe over denominator B apostrophe P end fraction

rightwards double arrow space Magnification comma space fraction numerator A apostrophe B apostrophe over denominator A B end fraction equals fraction numerator B apostrophe P over denominator B P end fraction

straight i. straight e. comma space straight m space equals negative straight v over straight u
So comma space straight m space equals space minus straight v over straight u
Hence comma space proved. space

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