
Light is incident from a medium into the air at two possible angles of incidence $(1)\,20^\circ $ and $(2)\,40^\circ $. In the medium, the light travels $3\,cm$ in $0.2ns$, the ray will:
A. Suffer total internal reflection in both cases
B. Suffer total internal reflection in case $2$only
C. Have $100\% $transmission in case $1$
D. Have partially reflection and partial transmission in case $2$.
Answer
163.5k+ views
Hint: To answer this question, we need to use the basic formula of the refractive index to calculate the refractive index of a medium. Then using it we can determine the expression for the critical angle of incidence, and hence conclude our answer.
Formula used:
The formulae used for solving this question are given by:
$\mu = \dfrac{{{v_a}}}{{{v_m}}}$, Here $\mu $is the refractive index of a medium in which the velocity of light is equal to ${v_m}$and ${v_a}$ the velocity of light in a vacuum.
${i_C} = {\sin ^{ - 1}}\left( {\dfrac{1}{\mu }} \right)$, here ${i_c}$is the critical angle of incidence when it travels from a medium of refractive index $\mu $ to the air.
Complete Step-by-Step Explanation:
Here, we have given the light travels in the medium is $3 \times {10^{ - 2}}$,
So, we have the velocity of the light in the medium is:
${v_m} = \dfrac{{3 \times {{10}^{ - 2}}}}{{0.2 \times {{10}^{ - 9}}}} \\$
$\Rightarrow {v_m} = 1.5 \times {10^8}\,m/s \\$
As we know, $\mu = \dfrac{{{v_a}}}{{{v_m}}}$
Substitute the obtained value in the above formula, then we have:
$\mu = \dfrac{{3 \times {{10}^8}}}{{1.5 \times {{10}^8}}} \\$
$\Rightarrow \mu = 2 \\$
The maximum incident angle at which the light ray will be reflected rather than refracted by the denser medium is known as the critical angle of incidence, so we have:
$\sin {\theta _{iC}} = \dfrac{{{n_2}}}{{{n_1}}}$
Therefore, the critical angle is: ${\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = 30^\circ $
[because refractive index of air is 1]
Now that both of the stated angles of incidence are below critical, refraction will occur and the beam gets partially absorbed during refraction.
Thus, the correct option is: (D) Have partially reflection and partial transmission in case $2$
Note: Remember the concept of critical angle. Light reflects back in the same medium only when it is traveling from denser to rarer medium and the angle of incidence becomes greater than the angle of refraction. However when a light ray strikes any surface, some part of it always gets absorbed by the medium in practical conditions.
Formula used:
The formulae used for solving this question are given by:
$\mu = \dfrac{{{v_a}}}{{{v_m}}}$, Here $\mu $is the refractive index of a medium in which the velocity of light is equal to ${v_m}$and ${v_a}$ the velocity of light in a vacuum.
${i_C} = {\sin ^{ - 1}}\left( {\dfrac{1}{\mu }} \right)$, here ${i_c}$is the critical angle of incidence when it travels from a medium of refractive index $\mu $ to the air.
Complete Step-by-Step Explanation:
Here, we have given the light travels in the medium is $3 \times {10^{ - 2}}$,
So, we have the velocity of the light in the medium is:
${v_m} = \dfrac{{3 \times {{10}^{ - 2}}}}{{0.2 \times {{10}^{ - 9}}}} \\$
$\Rightarrow {v_m} = 1.5 \times {10^8}\,m/s \\$
As we know, $\mu = \dfrac{{{v_a}}}{{{v_m}}}$
Substitute the obtained value in the above formula, then we have:
$\mu = \dfrac{{3 \times {{10}^8}}}{{1.5 \times {{10}^8}}} \\$
$\Rightarrow \mu = 2 \\$
The maximum incident angle at which the light ray will be reflected rather than refracted by the denser medium is known as the critical angle of incidence, so we have:
$\sin {\theta _{iC}} = \dfrac{{{n_2}}}{{{n_1}}}$
Therefore, the critical angle is: ${\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = 30^\circ $
[because refractive index of air is 1]
Now that both of the stated angles of incidence are below critical, refraction will occur and the beam gets partially absorbed during refraction.
Thus, the correct option is: (D) Have partially reflection and partial transmission in case $2$
Note: Remember the concept of critical angle. Light reflects back in the same medium only when it is traveling from denser to rarer medium and the angle of incidence becomes greater than the angle of refraction. However when a light ray strikes any surface, some part of it always gets absorbed by the medium in practical conditions.
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