Refraction by Prism - NEET Physics Chapterwise MCQs & PYQs
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NEET Refraction by Prism MCQs & PYQs
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Question 31:
easy
A thin prism having refracting angle $10^\circ$ is made of glass of refractive index $1.42$. This prism is combined with another thin prism of glass of refractive index $1.7$. This combination produces dispersion without deviation. The refracting angle of second prism should be:
(2017-Delhi)
For dispersion without deviation, the net deviation produced by the combination must be zero.
$(\mu_1 - 1)A_1 = (\mu_2 - 1)A_2$, where A and $\mu$ are refracting angle and refractive index.
$(1.42 - 1) \times 10^\circ = (1.7 - 1) \times A_2$, giving $4.2 = 0.7 A_2$, so $A_2 = 6^\circ$.
A ray is incident at an angle of incidence i on one surface of a small angle prism (with angle of prism A) and emerges normally from the opposite surface. If the refractive index of the material of the prism is m, then the angle of incidence is nearly equal to :
(2020)
Since the ray emerges normally, the angle of emergence $e = 0$, which implies $r_2 = 0$. For a prism, $r_1 + r_2 = A$, so $r_1 = A$. Using Snell's law for small angles, $i = \mu r_1$, which gives $i = \mu A$.
The refractive index of the material of a prism is $\sqrt{2}$ and the angle of the prism is $30^\circ$. One of the two refracting surfaces of the prism is made a mirror inwards, by silver coating. A beam of monochromatic light entering the prism from the other face will retrace its path (after reflection from the silvered surface) if its angle of incidence on the prism is:
(2018)
To retrace its path, the ray must strike the silvered surface normally, meaning $r_2 = 0$. From $r_1 + r_2 = A$, we get $r_1 = A = 30^\circ$. Applying Snell's law at the first surface: $1 \cdot \sin(i) = \mu \sin(r_1) = \sqrt{2} \sin(30^\circ) = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$. Therefore, the angle of incidence is $i = 45^\circ$.
The angle of incidence for a ray of light at a refracting surface of a prism is $45^\circ$. The angle of prism is $60^\circ$. If the ray suffers minimum deviation through the prism, the angle of minimum deviation and refractive index of the material of the prism respectively, are:
(2016 – I)
At minimum deviation, angle of emergence equals angle of incidence, so $i = e = 45^\circ$. Minimum deviation $\delta_m = i + e - A = 45^\circ + 45^\circ - 60^\circ = 30^\circ$. Refractive index $\mu = \frac{\sin((A+\delta_m)/2)}{\sin(A/2)} = \frac{\sin(45^\circ)}{\sin(30^\circ)} = \frac{1/\sqrt{2}}{1/2} = \sqrt{2}$.
The refracting angle of a prism is A, and refractive index of the material of the prism is $\cot(A/2)$. The angle of minimum deviation is:
(2015)
Refractive index $\mu = \frac{\sin((A+\delta_m)/2)}{\sin(A/2)}$. Given $\mu = \cot(A/2) = \frac{\cos(A/2)}{\sin(A/2)} = \frac{\sin(90^\circ - A/2)}{\sin(A/2)}$. Equating the numerators, we get $\frac{A + \delta_m}{2} = 90^\circ - \frac{A}{2}$. This simplifies to $A + \delta_m = 180^\circ - A$, so the angle of minimum deviation is $\delta_m = 180^\circ - 2A$.
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:
(2014)
For the ray to retrace its path, it must strike the silvered surface normally ($r_2 = 0$), so $r_1 = A$. Using Snell's law at the first surface: $\sin(i) = \mu \sin(r_1)$. Substituting $i = 2A$ and $r_1 = A$, we get $\sin(2A) = \mu \sin(A)$. Since $\sin(2A) = 2\sin(A)\cos(A)$, we find $\mu = 2\cos A$.
For the angle of minimum deviation of a prism to be equal to its refracting angle, the prism must be made of a material whose refractive index:
(2012 Mains)
Given $\delta_m = A$. The refractive index is $\mu = \frac{\sin((A+A)/2)}{\sin(A/2)} = \frac{\sin A}{\sin(A/2)} = 2\cos(A/2)$. For a physical prism, the maximum angle of incidence is $90^\circ$, which corresponds to $A = 90^\circ$ for $\delta_m = A$, giving $\mu = 2\cos(45^\circ) = \sqrt{2}$. As $A \to 0$, $\mu \to 2$, so the refractive index lies between $2$ and $\sqrt{2}$.
A ray of light is incident at an angle of incidence, i, on one face of prism of angle A (assumed to be small) and emerges normally from the opposite face. If the refractive index of the prism is $\mu$, the angle of incidence i, is nearly equal to:
(2012 Pre)
Since the ray emerges normally, $r_2 = 0$, which means $r_1 = A$. For small angles, Snell's law gives $i = \mu r_1$. Substituting $r_1 = A$, the angle of incidence is $i = \mu A$.
A thin prism of angle $15^\circ$ made of glass of refractive index $\mu_1 = 1.5$ is combined with another prism of glass of refractive index ( $\mu_2 = 1.75$ ). The combination of the prisms produces dispersion without deviation. The angle of the second prism should be:
(2011 Mains)
For dispersion without deviation, the net deviation is zero, so $(\mu_1 - 1)A_1 = (\mu_2 - 1)A_2$. Substituting the given values: $(1.5 - 1)(15^\circ) = (1.75 - 1)A_2$. This gives $0.5 \times 15^\circ = 0.75 A_2$, which simplifies to $7.5^\circ = 0.75 A_2$, so $A_2 = 10^\circ$.
A ray of light is incident on a $60^\circ$ prism at the minimum deviation position. The angle of refraction at the first face (incident face) of the prism is:
(2010 Mains)
At the minimum deviation position, the refracted ray is parallel to the base, and $r_1 = r_2 = A/2$. Given the prism angle $A = 60^\circ$, the angle of refraction at the first face is $r_1 = 60^\circ / 2 = 30^\circ$.