Solveeit Logo

Question

Question: Refractive index of water and glass are \[\dfrac{4}{3}\] and \[\dfrac{3}{2}\] respectively. A ray of...

Refractive index of water and glass are 43\dfrac{4}{3} and 32\dfrac{3}{2} respectively. A ray of light travelling in water is incident on the water-glass interface at 3030^\circ . Calculate the sine of the angle of refraction.
A. 0.4600.460
B. 0.5850.585
C. 0.4440.444
D. 0.6230.623

Explanation

Solution

You can start by giving a brief definition of Snell’s law and also write the equation of Snell’s law μ2μ1=sinisinr\dfrac{{{\mu _2}}}{{{\mu _1}}} = \dfrac{{\sin i}}{{\sin r}} . Then put all the corresponding values in the equation and calculate the value of sinr\sin r .

Complete answer:
Snell’s law defines a formula that establishes a relation between the angle of incidence and the angle of refraction, for a ray of light travelling from one medium to the other. Snell’s law is based on the fact that light travels with different speeds in different mediums and shows a bending effect when it travels from one medium to the other.
In this problem we are given a ray of light travelling from water to glass, the angle at which the ray of light strikes the water-glass interface is 3030^\circ . We are also given that the refractive index of water and glass are 43\dfrac{4}{3} and 32\dfrac{3}{2} respectively.

According to Snell’s law
μ2μ1=sinisinr\dfrac{{{\mu _2}}}{{{\mu _1}}} = \dfrac{{\sin i}}{{\sin r}}
Here, μ2={\mu _2} = The refractive index of the medium that ray of light travels to.
μ1={\mu _1} = The refractive index of the medium that ray of light travels from
i=i = The angle of incidence
r=r = The angle of incidence

So, given μ2=32={\mu _2} = \dfrac{3}{2} = Refractive index of glass
μ1=32={\mu _1} = \dfrac{3}{2} = Refractive index of water
And i=30i = 30
So, for this particular problem, the equation of Snell’s law becomes
3243=sin30sinr\dfrac{{\dfrac{3}{2}}}{{\dfrac{4}{3}}} = \dfrac{{\sin 30^\circ }}{{\sin r}}
3×32×4=12sinr\Rightarrow \dfrac{{3 \times 3}}{{2 \times 4}} = \dfrac{{\dfrac{1}{2}}}{{\sin r}}
sinr=49\Rightarrow \sin r = \dfrac{4}{9}
sinr=0.444\Rightarrow \sin r = 0.444

So, when the ray of light is incident on the water-glass interface at 3030^\circ , the angle that the refracted ray makes with the normal is 0.4440.444.

So, the correct answer is “Option C”.

Note:
In the given problem we are required to calculate the value of sinr\sin r . But you may also face problems where you have to calculate the value of rr . Most of the time the values are very simple and common trigonometric function like 3030^\circ , 6060^\circ , etc. but sometimes like in this problem you have to use the trigonometric table to find the answer (in this case r=26.8r = 26.8^\circ ).