
The refractive index for a prism is given as $\mu = \cot \dfrac{A}{2}$ then, angle of minimum deviation in terms of angle of prism is
A. $90^\circ - A$
B. $2A$
C. $180^\circ - A$
D. $180^\circ - 2A$
Answer
510.9k+ views
Hint: In order to solve this question we need to understand the definition of prism and scattering. Prism is a solid object with parallel and equivalents in shape and size, and with sides whose opposite edges are equal and parallel. Its work is to scatter white light in seven different colors. Scattering refers to splitting of white light in seven different colors as all seven colors have different wavelengths so different refractive index consequently different speed so they scatter.
Complete step by step answer:
Let the angle of prism be $A$ ,refractive index be $\mu $ and angle of minimum deviation be $m$.We know for prism the relation between refractive index, angle of prism and angle of minimum deviation is given by,
$\mu = \dfrac{{\sin (\dfrac{{m + A}}{2})}}{{\sin (\dfrac{A}{2})}}$
According to the Question $\mu = \cot (\dfrac{A}{2})$
Putting values we get
$\cot (\dfrac{A}{2}) = \dfrac{{\sin (\dfrac{{m + A}}{2})}}{{\sin (\dfrac{A}{2})}}$
Since, we know that
$\cot (\dfrac{A}{2}) = \dfrac{{\cos (\dfrac{A}{2})}}{{\sin (\dfrac{A}{2})}}$
Hence we get,
$\dfrac{{\cos (\dfrac{A}{2})}}{{\sin (\dfrac{A}{2})}} = \dfrac{{\sin (\dfrac{{m + A}}{2})}}{{\sin (\dfrac{A}{2})}}$
$ \Rightarrow \cos (\dfrac{A}{2}) = \sin (\dfrac{{m + A}}{2})$
Since we know, $\cos (\dfrac{A}{2}) = \sin (90^\circ - \dfrac{A}{2})$ so,
$\sin (90^\circ - \dfrac{A}{2}) = \sin (\dfrac{{m + A}}{2})$
$ \Rightarrow 90^\circ - \dfrac{A}{2} = \dfrac{{m + A}}{2}$
$\dfrac{{m + A}}{2} = \dfrac{{180^\circ - A}}{2}$
$ \Rightarrow m = 180^\circ - A - A$
$\therefore m = 180^\circ - 2A$
So the correct option is D.
Note: It should be remembered that we are calculating mean deviation of a single ray out of seven different colors which is represented by VIBGYOR expands as Violet, Indigo, Blue, Green ,Yellow, Orange and Red. It also should be noted that natural prisms are rain droplets which cause a rainbow in the sky. Also red in the sky scatters least and blue color scatters more because of wavelength as red has the largest wavelength.
Complete step by step answer:
Let the angle of prism be $A$ ,refractive index be $\mu $ and angle of minimum deviation be $m$.We know for prism the relation between refractive index, angle of prism and angle of minimum deviation is given by,
$\mu = \dfrac{{\sin (\dfrac{{m + A}}{2})}}{{\sin (\dfrac{A}{2})}}$
According to the Question $\mu = \cot (\dfrac{A}{2})$
Putting values we get
$\cot (\dfrac{A}{2}) = \dfrac{{\sin (\dfrac{{m + A}}{2})}}{{\sin (\dfrac{A}{2})}}$
Since, we know that
$\cot (\dfrac{A}{2}) = \dfrac{{\cos (\dfrac{A}{2})}}{{\sin (\dfrac{A}{2})}}$
Hence we get,
$\dfrac{{\cos (\dfrac{A}{2})}}{{\sin (\dfrac{A}{2})}} = \dfrac{{\sin (\dfrac{{m + A}}{2})}}{{\sin (\dfrac{A}{2})}}$
$ \Rightarrow \cos (\dfrac{A}{2}) = \sin (\dfrac{{m + A}}{2})$
Since we know, $\cos (\dfrac{A}{2}) = \sin (90^\circ - \dfrac{A}{2})$ so,
$\sin (90^\circ - \dfrac{A}{2}) = \sin (\dfrac{{m + A}}{2})$
$ \Rightarrow 90^\circ - \dfrac{A}{2} = \dfrac{{m + A}}{2}$
$\dfrac{{m + A}}{2} = \dfrac{{180^\circ - A}}{2}$
$ \Rightarrow m = 180^\circ - A - A$
$\therefore m = 180^\circ - 2A$
So the correct option is D.
Note: It should be remembered that we are calculating mean deviation of a single ray out of seven different colors which is represented by VIBGYOR expands as Violet, Indigo, Blue, Green ,Yellow, Orange and Red. It also should be noted that natural prisms are rain droplets which cause a rainbow in the sky. Also red in the sky scatters least and blue color scatters more because of wavelength as red has the largest wavelength.
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