
A parallel sides glass slab of thickness 4 cm is made of a material of refractive index $\sqrt 3 $ When light is incident on one of the parallel faces at an angle of 60, it emerges from the other parallel face. Then the lateral displacement of the emergent beam.
A. $\dfrac{4}{{\sqrt 3 }}$
B. $4\sqrt 3 $
C. $\dfrac{3}{2}$
D. $\dfrac{3}{{\sqrt 2 }}$
Answer
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Hint: First of all we will define lateral displacement and explain it. Write the formula of lateral displacement and then relate the term. In refraction, emergent ray is parallel to the incident ray but in actual it there is slightly shifted and this shift in the position of the emergent ray and the incident ray is known as lateral displacement.
Formula used:
Lateral displacement can be calculated using:
$S = \dfrac{t}{{\cos r}}\sin (i - r)$
Where S = lateral shift t = thickness of the medium i = angle of incidence r = angle of refraction
Complete answer:
Lateral displacement is the perpendicular distance between the incident and emergent ray. The formula given by,
$S = \dfrac{t}{{\cos r}}\sin (i - r)$
Where S = lateral shift t = thickness of the medium i = angle of incidence r = angle of refraction
So, the lateral displacement depends upon the thickness of the medium. the angle of incidence, the angle of refraction.
Snell’s law states that the ratio of the sin angle of incidence to the sine angle of refraction gives the refractive index of the medium.
$\mu = \dfrac{{\sin i}}{{\sin r}}$
Substituting the value of $\mu $and$i$we get,
$\sqrt 3 = \dfrac{{\sin 60}}{{\sin r}}$
$ \Rightarrow \sin r = \dfrac{{\sin 60}}{{\sqrt 3 }} = \dfrac{1}{2}$
$r = 30 \circ $
Putting this in the lateral displacement formula we get
$S = \dfrac{4}{{\cos 30}}\sin (60 - 30)$
$ \Rightarrow S = \dfrac{4}{{\cos 30}}\sin (30)$
$ \Rightarrow S = \dfrac{4}{{\sqrt 3 }}$
Therefore, option (A) is the correct answer.
Note: Students should take care about the factors affecting the lateral displacement by the formula given .Refractive Index is defined as the ratio of the speed of light in a vacuum to the speed of light in a second medium of having more density. The refractive index depends upon the wavelength of beam, this causes white light to split into constituent colours on refraction. This is called dispersion. The term normal dispersion is defined as the refractive index is higher for blue coloured light than for red light colour.
Formula used:
Lateral displacement can be calculated using:
$S = \dfrac{t}{{\cos r}}\sin (i - r)$
Where S = lateral shift t = thickness of the medium i = angle of incidence r = angle of refraction
Complete answer:
Lateral displacement is the perpendicular distance between the incident and emergent ray. The formula given by,
$S = \dfrac{t}{{\cos r}}\sin (i - r)$
Where S = lateral shift t = thickness of the medium i = angle of incidence r = angle of refraction
So, the lateral displacement depends upon the thickness of the medium. the angle of incidence, the angle of refraction.
Snell’s law states that the ratio of the sin angle of incidence to the sine angle of refraction gives the refractive index of the medium.
$\mu = \dfrac{{\sin i}}{{\sin r}}$
Substituting the value of $\mu $and$i$we get,
$\sqrt 3 = \dfrac{{\sin 60}}{{\sin r}}$
$ \Rightarrow \sin r = \dfrac{{\sin 60}}{{\sqrt 3 }} = \dfrac{1}{2}$
$r = 30 \circ $
Putting this in the lateral displacement formula we get
$S = \dfrac{4}{{\cos 30}}\sin (60 - 30)$
$ \Rightarrow S = \dfrac{4}{{\cos 30}}\sin (30)$
$ \Rightarrow S = \dfrac{4}{{\sqrt 3 }}$
Therefore, option (A) is the correct answer.
Note: Students should take care about the factors affecting the lateral displacement by the formula given .Refractive Index is defined as the ratio of the speed of light in a vacuum to the speed of light in a second medium of having more density. The refractive index depends upon the wavelength of beam, this causes white light to split into constituent colours on refraction. This is called dispersion. The term normal dispersion is defined as the refractive index is higher for blue coloured light than for red light colour.
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