The angle of a prism is A. One of its refracting surfaces is silvered. Light rays falling at an angle of incidence 2A on the first surface returns back through the same path after suffering reflection at the silvered surface.The refractive index \\(\mu\\), of the prism is :

To retrace its path light rays should fall normally on the reflecting surface. So\\(r_2= 0\\)\\(\Rightarrow r_1 =A- r_2 \Rightarrow r_1 = A\\)Now applying snell rule between incident ray and refracted ray.\\((1) sin (2A) = n sin (A)\Rightarrow 2 sin A cos A= n sin A\\)\\(\Rightarrow n = 2 cos A\\)

How does the angle of minimum deviation of a glass prism vary, if the incident violet light is replaced by red light? Give reason.

Refractive index is inversely proportional to the wavelength of the light. Red has a higher wavelength than violet light therefore it has a smaller refractive index than violet.Again angle of minimum deviation is directly proportional to refractive index and hence will decrease when replaced by red light.

A glass prism of refractive index \\(1.5\\) is immersed in water (refractive index \\(4/3)\\). A light beam normally on the face \\(AB\\) is totally reflected to reach on the face \\(BC\\) if

A light beam incident normally on the face AB is totally reflected to reach on the face BC if \\(\theta\geq\theta_c\\)The beam is grazed on face BC, \\(e=90^o\\)Refractive index of water,\\(n_1=\dfrac{4}{3}\\) and that of prism,\\(n_2=1.5\\)using Snell's law of refraction, \\(n_2sin \theta_c=n_1sine\\)\\(sin\theta_c=\dfrac{8}{9}\\)For total internal reflection \\(sin\theta\geq sin\theta_c\\)Hence \\(sin\theta \geq \dfrac{8}{9}\\)

A light ray is incident perpendicularly to one face of a 90\\(^{\circ}\\) prism and is totally internally reflected at the glass-air interface. If the angle of reflection is 45\\(^{\circ}\\), we conclude that the refractive index \\(n\\) :

We know\\(\theta _{cric}=sin^{-1}(\dfrac{1}{n })\\)\\(45^{o}> \theta \\)\\(sin(45^o)> \dfrac{1}{n}\\) \\(\Rightarrow n > \sqrt{2} \\)

When light rays are incident on a prism at an angle of \\(45^o\\), the minimum deviation is obtained. If the refractive index of the material of the prism is \\(\sqrt 2\\), then the angle of prism will be

\\(\begin{array}{l} Let\, \, \angle i\, \, \, denote\, \, incidence\, \, angle\, \, and\, \angle r\, \, denote\, \, refracted\, \, angle. \\ Given: \\ \angle i=2\times \angle r \\ Now\, \, by\, \, snell's\, \, Law\, \, appling\, \, at\, \, { { int } }erface\, \, of\, \, ism: \\ \frac { { \sin i } }{ { \sin r } } =\frac { { \sin \left( { 2r } \right) } }{ { \sin r } } =\frac { { 2\sin r\times \cos r } }{ { \sin r } } =2\cos r=\sqrt { 2 } \\ where\, \, we\, \, have\, \, used\, \, trignometric\, \, identity; \\ \sin 2r=2\sin x\times \cos x \\ Therefore\, \, we\, \, have; \\ \cos r=\frac { { \sqrt { 2 } } }{ 2 } =\frac { 1 }{ { \sqrt { 2 } } } \\ \Rightarrow r={ 45^{ 0 } } \\ now\, \, we\, \, have3\, \, relation\, \, from\, \, a\, \, \min imum\, \, deviation; \\ A={ r_{ 1 } }+{ r_{ 2 } } \\ A=2r \\ A={ 90^{ 0 } } \end{array}\\)

Light is incident normally on face AB of a prism as shown in figure. A liquid of refractive index \\(\mu\\) is placed on the face AC of the prism. The prism is made of glass of refractive index 3/2. The limit of \\(\mu\\) for which total internal reflection takes place on face AC is :

Critical angle for glass-liquid : \\(_a\mu_g\, sin \, C=_a\mu _i\, sin 90^o\\)\\(\Rightarrow \dfrac{3}{2} sin \, C=\mu \times 1\\)\\(\Rightarrow sin \, C =\dfrac{2\mu}{3}\\)For Total Internal reflection on AC,We must have \\(60^o \geq C\\)\\(\Rightarrow sin 60^o \geq sin \, C\\)\\(\Rightarrow \dfrac{\sqrt{3}}{2}\geq \dfrac{2 \mu}{3}\\)Thus, we get \\(\mu\leq \dfrac{3\sqrt{3}}{4}\\)

The refractive indices of violet and red light are \\(1.54\\) and \\(1.52\\) respectively. If the angle prism is \\(100\\) , then the angular dispersion is :

Angular dispersion is given as, \\(\alpha = \left( {{\mu _V} - {\mu _R}} \right)A\\) \\(\alpha = \left( {1.54 - 1.52} \right) \times 100\\) \\(\alpha = 0.02 \times 100\\) \\(\alpha = 2\\) Thus, angular dispersion is \\(2\;^\circ \\).

If one face of a prism of prism angle \\(30^o\\) and \\(\mu\sqrt{2}\\) is silvered, the incident ray retraces its intial path. The angle of incidence is:

For incident light ray to retrace it's initial path, it should fall normally on the silvered surface.Given, prism angle, \\( A = 30{^{\circ}} \\)Let the angle of incidence be \\( i \\)From the figure, \\( x = 90 - A = 60{^{\circ}} \\)\\( x + r = 90{^{\circ}} \Rightarrow \angle r = 30{^{\circ}} \\)Thus, angle of refraction \\( \angle r = 30{^{\circ}} \\)Now, applying Snell's law at the surface DE, we have\\( \mu = \dfrac{Sin\ i}{Sin\ r} \\)\\( Sin\ i = Sin\ r \times \mu \\)\\( Sin\ i = Sin\ 30 \times \sqrt{2} \\)\\( i = Sin^{-1}(\dfrac{\sqrt{2}}{2} ) \\)\\( i = Sin^{-1}(\dfrac{1}{\sqrt{2}} ) \\)\\( i = 45{^{\circ}} \\)Hence, the answer is OPTION C.

Problems on prism formula - L1 | Definition, Examples, Diagrams

Q: At what angle should a ray of light be incident on the face of a prism of refracting angle \\(60^\circ\\) so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is \\(1.524\\).

Angle of prism A is \\(60^o; \mu=1.524\\)\\(i\\) is angle of incidence.\\(r_1\\) is angle of refraction.\\(r_2\\) is angle of incidence during exit from other end.\\(e\\) is 90^o angle of convergent.\\(\mu=sine/sinr_2\Rightarrow r_2=41^o\\) \\(r_1=A-r_2=19^o\\)\\(\Rightarrow \mu=sini/sinr_1\Rightarrow i=29.75^o\\)

example

Problems on Total internal reflection in Prism

Example: At what angle should a ray of light be incident on the face of a prism of refracting angle

60^{\circ}

so that it just suffers total internal reflection at the other face. Find the refractive index of the prism.Refractive index of prism is 1.524.

Solution:

sin c = \frac{1}{μ} = \frac{1}{1 . 5 2 4} = 0.65 o r c = 40^{\circ} 30^{^{'}} r = 19^{\circ} 30^{^{'}} = [180^{\circ} - 120^{\circ} - 40^{\circ} 30^{^{'}}] μ = \frac{sin i}{sin r} o r sin i = 1.524 s i n 19^{\circ} 30^{^{'}} = 1.524 (. 3340) = 0.5 o r i = 30^{\circ}

example

Solve problems on light ray passing through multiple prisms

Example:
A ray of light enters at grazing angle of incidence into an assembly of three isosceles right-angled prisms having refractive indices

μ_{1} = 2, μ_{2} = x

and

μ_{3} = 3

. If finally emergent light ray also emerges at grazing angle then calculate x.

Solution:
Apply Snell's law on various surfaces one by one :
1 sin

9 0^{\circ}

= μ_{1}

sin

r_{1} \Rightarrow

sin

r_{1} = \frac{1}{2} \Rightarrow r_{1} = 4 5^{\circ}

We have the incident angle for the second interface given using the relation sin

i_{2}

=cos

r_{1}

Thus we get

μ_{1}

cos

r_{1}

μ_{2}

sin

r_{2} \Rightarrow

sin

r_{2}

\frac{1}{μ _{2}}

(cos

r_{2}

\frac{1}{2}

)

μ_{2}

cos

r_{2} = μ_{3}

sin

r_{3}

\Rightarrow

sin

r_{3}

= \frac{μ _{2} 1 - s i n ^{2} r _{2}}{3}

μ_{3}

cos

r_{3} = 1

\frac{μ _{2}^{2} - 1}{3}

sin

^{2} r_{3}

+ cos

^{2} r_{3} = 1

\Rightarrow \frac{μ _{2}^{2} - 1}{3} + \frac{1}{3} = 1 \Rightarrow μ_{2}^{2} = 3 \Rightarrow μ_{2} = 3

Thus we get

x = 3 .

Problems on prism formula - L1

example

Problems on Total internal reflection in Prism

example

Solve problems on light ray passing through multiple prisms

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