Question

A ray of light strikes a transparent surface from air at an angle θ. If the angle between the reflected and refrected ray is a right angle,the refreactive index of the other surface is given by.

Answer 1

Answer:

Let ' I ' be the angle of incidence and reflection and 'r' be the angle of refraction in the optical medium.

As per the given condition, the angle between reflected and refracted rays is 90° . hence we have

I + r + 90° = 180°

r = 90° - I

If n is the refractive index of optical medium air these using Snell's law as follows

sin I = n sin r

sin I = n sin (90° - i)

sin I = n cos I

n = tan I

Where 'I ' is the angle of incidence on transparent ( optical) surface which should known .

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Answer 2

Answer:

The refractive index of the other surface is tanθ.

Explanation:

Given angle of incident of light ray on transparent surface, $i=\theta$

The angle between the reflected and refracted ray is 90°.

From laws of reflection, If θ is the angle of incidence, then the angle of reflection(r) is also θ, hence $r=\theta$.

Let $t$ be the angle of refraction.

Given that the angle between reflected and refracted rays is 90°, i.e.,

$t + \theta = 90^o$

$\implies t = 90^o- \theta$

According to laws of refraction, the ratio of the sin of the angle of incidence to the sin of the angle of refraction is a constant, called refractive index($\mu$). This law is also called Snell's law of refraction.

So, according to Snell's law,

$\dfrac{sin~ i}{sin~t} = \mu$

Substituting $i$ and $t$,

$\dfrac{sin~ \theta}{sin~ ( 90^o- \theta)} = \mu$

Since $sin~ ( 90^o- \theta)} =cos~ \theta}$

$\implies \dfrac{sin~ \theta}{cos~ \theta} = \mu$

$\implies tan~ \theta = \mu$

Therefore, the refractive index of the other surface is tanθ.