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Hint: Whenever a ray travels from denser to rarer or rarer to denser medium there is partial reflection and refraction of light. And, the angle of incidence is always equal to the angle of reflection. Using all this information we will solve this question. The snell’s law is also to be used here.
Complete step by step solution:
Refractive index is given $\mu = 1.5$. Also we know that the angle of incidence is equal to the angle of reflection.
$\angle i = \angle r$
We have Snell’s law as below.
$\dfrac{{\sin i}}{{\sin r'}} = \mu $
Here, $i$ is the angle of incidence, $r'$ is the angle of refraction, and $\mu $ is the refractive index of glass.
Let us substitute the values in the above equation.
$\dfrac{{\sin i}}{{\sin r'}} = 1.5$
We can also replace the angle of incidence with the angle of reflection in the above formula.
$\dfrac{{\sin r}}{{\sin r'}} = 1.5$ ………………….(1)
Now, it is given in the question that the reflected and refracted angles are at a right angle to each other.
$\angle r + \angle r' = 90$
$ \Rightarrow \angle r' = 90 - \angle r$
Put this value in equation (1) and we get the following.
$\dfrac{{\sin r}}{{\sin \left( {90 - r} \right)}} = 1.5$
$ \Rightarrow \dfrac{{\sin r}}{{\cos r}} = 1.5$
We can further write it as below.
$\tan r = 1.5$
$ \Rightarrow r = {\tan ^{ - 1}}\dfrac{3}{2} = {\sin ^{ - 1}}\left( {\dfrac{{3/2}}{{\sqrt {1 + {{\left( {3/2} \right)}^2}} }}} \right) = {\sin ^{ - 1}}\dfrac{3}{{\sqrt {13} }}$
Hence, the correct option is (D) ${\sin ^{ - 1}}\dfrac{3}{{\sqrt {13} }}$.
Note:
The ratio of the sine of incidence to the sine of refraction is constant for the given pair of the medium. This is called Snell’s law.
When a ray of light goes from rarer to denser it bends towards the normal and when a ray of light goes from denser to rarer, the ray of light moves away from the normal.
Complete step by step solution:
Refractive index is given $\mu = 1.5$. Also we know that the angle of incidence is equal to the angle of reflection.
$\angle i = \angle r$
We have Snell’s law as below.
$\dfrac{{\sin i}}{{\sin r'}} = \mu $
Here, $i$ is the angle of incidence, $r'$ is the angle of refraction, and $\mu $ is the refractive index of glass.
Let us substitute the values in the above equation.
$\dfrac{{\sin i}}{{\sin r'}} = 1.5$
We can also replace the angle of incidence with the angle of reflection in the above formula.
$\dfrac{{\sin r}}{{\sin r'}} = 1.5$ ………………….(1)
Now, it is given in the question that the reflected and refracted angles are at a right angle to each other.
$\angle r + \angle r' = 90$
$ \Rightarrow \angle r' = 90 - \angle r$
Put this value in equation (1) and we get the following.
$\dfrac{{\sin r}}{{\sin \left( {90 - r} \right)}} = 1.5$
$ \Rightarrow \dfrac{{\sin r}}{{\cos r}} = 1.5$
We can further write it as below.
$\tan r = 1.5$
$ \Rightarrow r = {\tan ^{ - 1}}\dfrac{3}{2} = {\sin ^{ - 1}}\left( {\dfrac{{3/2}}{{\sqrt {1 + {{\left( {3/2} \right)}^2}} }}} \right) = {\sin ^{ - 1}}\dfrac{3}{{\sqrt {13} }}$
Hence, the correct option is (D) ${\sin ^{ - 1}}\dfrac{3}{{\sqrt {13} }}$.
Note:
The ratio of the sine of incidence to the sine of refraction is constant for the given pair of the medium. This is called Snell’s law.
When a ray of light goes from rarer to denser it bends towards the normal and when a ray of light goes from denser to rarer, the ray of light moves away from the normal.
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