Question

In refraction incident ray and emergent ray are parallel. Type 1 for true and 0 for false.

Answer 1

Hint: To answer this question, we need to take the case of refraction through a glass slab. Then we need to examine whether the emergent ray is parallel to the incident ray, using Snell's law.
Formula Used: The formula used in this solution is
\[\mu = \dfrac{{\sin i}}{{\sin r}}\], \[\mu = \]refractive index, \[i = \]angle of incidence, and \[r = \] angle of refraction

Complete step-by-step answer
Consider a glass slab of refractive index \[\mu \] as shown in the flowing figure.
A ray \[AB\] is incident on the upper interface at the point \[C\], which gets refracted in the glass medium as the ray \[CD\]. The refracted ray is incident on the lower interface at the point \[E\]. Finally, \[CD\] is refracted and emerges out of the glass medium back into the air as the ray\[EF\].
Applying Snell’s law on the upper interface, we get
\[\mu = \dfrac{{\sin i}}{{\sin r}}\] (i)
Again applying Snell’s law on the lower interface, we get

\[\mu = \dfrac{{\sin e}}{{\sin r'}}\] (ii)
As \[{N_1}{N_2}\] is the normal to the upper interface,
\[\therefore {N_1}{N_2} \bot PQ\] (iii)
Also, \[{N’_1}{N’_2}\] is the normal to the lower surface
\[\therefore {N’_1}{N’_2} \bot RS\] (iv)
Case 1
If \[PQ\parallel RS\] then by (iii) and (iv), we have
\[{N_1}{N_2}\parallel {N’_1}{N’_2}\]
Now, in the figure, as \[r\] and \[r'\] are alternate interior angles, so they are equal.
\[\therefore r = r'\]
Taking sine both the sides, we get
\[\sin r = \sin r'\] (v)
From (i) and (ii)
\[\dfrac{{\sin i}}{{\sin r}} = \dfrac{{\sin e}}{{\sin r'}}\]
Multiplying with \[\sin r\] on both sides
\[\sin i = \sin r\dfrac{{\sin e}}{{\sin r'}}\]
From (v)
\[\sin i = \sin e\]
\[\because \] both \[i\& e\] are acute, so we can say
\[i = e\]
As \[{N_1}{N_2} \parallel {N’_1}{N’_2}\] and \[i = e\]
So \[AB\parallel EF\]
Hence, the incident ray and emergent ray are parallel.
Case 2
If \[PQ\& RS\] are not parallel, then
\[r \ne r'\]
both \[i\& e\] are acute, so we can say
\[\sin r \ne \sin r'\]
 From (i) and (ii)
\[\dfrac{{\sin i}}{{\sin r}} = \dfrac{{\sin e}}{{\sin r'}}\]
But \[\sin r \ne \sin r'\]
So, \[\sin i \ne \sin e\] and therefore
\[i \ne e\]
Thus, \[AB\] is not parallel to \[EF\]
\[\therefore \] We conclude that the incident and the emergent rays are parallel only with the condition that the two refracting interfaces should be parallel to each other. They are not always parallel.
So the statement given in the question is false.

Hence, the correct answer is 0.

Note: This question can also be attempted by taking the case of refraction through a triangular prism. In that case, we can directly conclude that the emergent ray is not parallel to the incident ray through simple observation.