
A ray of light travelling in a transparent medium falls on a surface separating the medium from air, at an angle of incidence of ${45^0}$. The ray undergoes total internal reflection. If n is the refractive index of the medium with respect to air, select the possible values of n from the following
(A) $1.3$
(B) $1.4$
(C) $1.5$
(D) $1.6$
Answer
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Hint
Here we will see that the condition of total internal reflection is satisfied and then use the basic formula of refractive index to calculate the value of n
$\Rightarrow \mu = \dfrac{{\sin i}}{{\sin r}}$
Complete step by step answer
It is given that angle of incidence, $i = {45^0}$
We know for total internal reflection, angle of incidence must be greater than critical angle of the two media,
Critical angle, $c = {\sin ^{ - 1}}\left( {\dfrac{1}{\mu }} \right)$ (since angle of redfraction $r = {90^0}$)
Condition for total internal reflection $i > {\sin ^{ - 1}}\left( {\dfrac{1}{\mu }} \right)$
Or, $\sin i > \left( {\dfrac{1}{\mu }} \right) \Rightarrow \mu > \dfrac{1}{{\sin i}}$
$\Rightarrow \mu > \dfrac{1}{{\sin {{45}^0}}}$
$\Rightarrow \mu > \sqrt 2 $
$\Rightarrow \mu > 1.44$
From the options we can see that refractive index greater than $1.44$ are $1.5$ and $1.6$. Hence, the possible values of n are $1.5$ and $1.6$.
The correct options are (C) $1.5$ and (D) $1.6$.
Additional Information
On travelling from denser to rarer medium ,a ray of light bends away from the normal at interface. Critical angle for two media is defined as that angle of incidence in the denser medium for which the rarer medium has an angle of refraction as ${90^0}$. The value of critical angle depends on the color of light and nature of the two media. Critical angles differ for various colours because of the dependence of the refractive index on the wavelength of ray.
Note
Total internal reflection is the phenomenon in which a light ray travelling from denser medium to a rarer medium which is incident at the interface of two media at an angle greater than critical angle, then the ray is totally reflected into the denser medium.Example- The brightness of diamond is because of its small critical angle which causes total internal reflection.
Here we will see that the condition of total internal reflection is satisfied and then use the basic formula of refractive index to calculate the value of n
$\Rightarrow \mu = \dfrac{{\sin i}}{{\sin r}}$
Complete step by step answer
It is given that angle of incidence, $i = {45^0}$
We know for total internal reflection, angle of incidence must be greater than critical angle of the two media,
Critical angle, $c = {\sin ^{ - 1}}\left( {\dfrac{1}{\mu }} \right)$ (since angle of redfraction $r = {90^0}$)
Condition for total internal reflection $i > {\sin ^{ - 1}}\left( {\dfrac{1}{\mu }} \right)$
Or, $\sin i > \left( {\dfrac{1}{\mu }} \right) \Rightarrow \mu > \dfrac{1}{{\sin i}}$
$\Rightarrow \mu > \dfrac{1}{{\sin {{45}^0}}}$
$\Rightarrow \mu > \sqrt 2 $
$\Rightarrow \mu > 1.44$
From the options we can see that refractive index greater than $1.44$ are $1.5$ and $1.6$. Hence, the possible values of n are $1.5$ and $1.6$.
The correct options are (C) $1.5$ and (D) $1.6$.
Additional Information
On travelling from denser to rarer medium ,a ray of light bends away from the normal at interface. Critical angle for two media is defined as that angle of incidence in the denser medium for which the rarer medium has an angle of refraction as ${90^0}$. The value of critical angle depends on the color of light and nature of the two media. Critical angles differ for various colours because of the dependence of the refractive index on the wavelength of ray.
Note
Total internal reflection is the phenomenon in which a light ray travelling from denser medium to a rarer medium which is incident at the interface of two media at an angle greater than critical angle, then the ray is totally reflected into the denser medium.Example- The brightness of diamond is because of its small critical angle which causes total internal reflection.
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