To ensure almost 100% transmittivity, photographic lenses are often coated with a thin layer of dielectric material. The refractive index of this material is intermediated between that of air and glass (which makes the optical element of the lens). A typically used dielectric film is MgF2 (n=138). What should the thickness of the film be so that at the centre of the visible spectrum (5500 Å) there is maximum transmission.

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- refractive index = 1.38 refractive index = 1.5

^{$\lambda =5500A$ 0}

Consider a ray incident at an angle i. A part of this ray is reflected from the air-film interface And apart refracted inside.

This is partly reflected at the film-glass interface and a part transmitted. A part of the

reflected ray is reflected at the film-air interface and a part transmitted as r2 parallel to r 1. Of course successive reflections and transmissions will keep on decreasing the amplitude of the wave. Hence, rays r 1 and r2 shall dominate the behaviour. If incident light is to be transmitted thro

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This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- refractive index = 1.38 refractive index = 1.5

^{$\lambda =5500A$ 0}

Consider a ray incident at an angle i. A part of this ray is reflected from the air-film interface And apart refracted inside.

This is partly reflected at the film-glass interface and a part transmitted. A part of the

reflected ray is reflected at the film-air interface and a part transmitted as r2 parallel to r 1. Of course successive reflections and transmissions will keep on decreasing the amplitude of the wave. Hence, rays r 1 and r2 shall dominate the behaviour. If incident light is to be transmitted through the lens, r 1 and r2 should interfere destructively. Both the reflections at A and D are from lower to higher refractive index and hence, there is no phase change on reflection. The optical path difference between r₂ and r₁ is

n(AD+CD)-AB

if d is the thickness of the film , then AD=CD=d/cosr

AB=AC sini

AC/2= d tanr

AC= 2dtanr

Hence AB= 2d tanr sini

Thus the optical path difference = $\frac{2nd}{cosr}$ -2dtanrsini

= 2. $\frac{sinid}{sinrcosr}$ - 2d $\frac{sinr}{cosr}sini$

= 2dsin{ $\frac{1-{sin}^{2}r}{sinrcosr}$ }

= 2ndcosr

For path difference $\frac{\lambda }{2}$

2ndcosr= $\frac{\lambda }{2}$

ndcosr= $\frac{\lambda }{4}$

for photographic lenses, the sources are normally in vertical plane

i=r=0⁰

ndcos0= $\frac{\lambda }{4}$

d= $\frac{\lambda }{4n}$ =5500/4 $\times 1.38$ = 1000A⁰

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<p>This is a Long Answer Type Questions as classified in NCERT Exemplar</p><p><strong>Explanation-</strong> refractive index = 1.38 refractive index = 1.5</p><p>^{<span contenteditable="false"> <math> <mi>λ</mi> <mo>=</mo> <mn>5500</mn> <mi>A</mi> </math> </span>0}</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1743139677phpsbOb6e_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1743139677phpsbOb6e.jpeg" alt="" width="311" height="224"></picture></div></div><p>Consider a ray incident at an angle i. A part of this ray is reflected from the air-film interface And apart refracted inside.</p><p>This is partly reflected at the film-glass interface and a part transmitted. A part of the</p><p>reflected ray is reflected at the film-air interface and a part transmitted as r2 parallel to r 1. Of course successive reflections and transmissions will keep on decreasing the amplitude of the wave. Hence, rays r 1 and r2 shall dominate the behaviour. If incident light is to be transmitted through the lens, r 1 and r2 should interfere destructively. Both the reflections at A and D are from lower to higher refractive index and hence, there is no phase change on reflection. The optical path difference between r₂ and r₁ is</p><p>n(AD+CD)-AB</p><p>if d is the thickness of the film , then AD=CD=d/cosr</p><p>AB=AC sini</p><p>AC/2= d tanr</p><p>AC= 2dtanr</p><p>Hence AB= 2d tanr sini</p><p>Thus the optical path difference = <span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mn>2</mn> <mi>n</mi> <mi>d</mi> </mrow> </mrow> <mrow> <mrow> <mi>c</mi> <mi>o</mi> <mi>s</mi> <mi>r</mi> </mrow> </mrow> </mfrac> </math> </span>-2dtanrsini</p><p>= 2.<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mi>i</mi> <mi> </mi> <mi>d</mi> </mrow> </mrow> <mrow> <mrow> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mi>r</mi> <mi>c</mi> <mi>o</mi> <mi>s</mi> <mi>r</mi> </mrow> </mrow> </mfrac> </math> </span>- 2d<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mi>c</mi> <mi>o</mi> <mi>s</mi> <mi>r</mi> </mrow> </mrow> </mfrac> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mi>i</mi> </math> </span></p><p>= 2dsin{<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mn>1</mn> <mo>-</mo> <msup> <mrow> <mrow> <mi>s</mi> <mi>i</mi> <mi>n</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mi>r</mi> <mi>c</mi> <mi>o</mi> <mi>s</mi> <mi>r</mi> </mrow> </mrow> </mfrac> </math> </span>}</p><p>= 2ndcosr</p><p>For path difference<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </math> </span></p><p>2ndcosr=<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </math> </span></p><p>ndcosr=<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </mfrac> </math> </span></p><p>for photographic lenses, the sources are normally in vertical plane</p><p>i=r=0⁰</p><p>ndcos0=<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </mfrac> </math> </span></p><p>d=<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> <mi>n</mi> </mrow> </mrow> </mfrac> </math> </span>=5500/4<span contenteditable="false"> <math> <mo>×</mo> <mn>1.38</mn> </math> </span>= 1000A⁰</p>

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