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Question: A ray of light is incident on a medium of refractive index '$\mu$' at an angle of incidence '$i$'. O...

A ray of light is incident on a medium of refractive index 'μ\mu' at an angle of incidence 'ii'. On refraction in the medium 'δ\delta' is the angle of deviation. Then

A

1μ=cos⁡δ−sin⁡δtan⁡i\frac{1}{\mu} = \cos \delta - \frac{\sin \delta}{\tan i}

B

1μ=sin⁡δ−cos⁡δtan⁡i\frac{1}{\mu} = \sin \delta - \frac{\cos \delta}{\tan i}

C

1μ=cos⁡δ−sin⁡δ⋅tan⁡i\frac{1}{\mu} = \cos \delta - \sin \delta \cdot \tan i

D

1μ=sin⁡δ−cos⁡δ⋅tan⁡i\frac{1}{\mu} = \sin \delta - \cos \delta \cdot \tan i

Answer

1μ=cos⁡δ−sin⁡δtan⁡i\frac{1}{\mu} = \cos \delta - \frac{\sin \delta}{\tan i}

Explanation

Solution

Given:

  • Incident angle: ii
  • Refractive index of medium: μ\mu
  • Angle of deviation: δ\delta

When a ray refracts, the deviation is defined as:

δ=i−r⇒r=i−δ\delta = i - r \quad \Rightarrow \quad r = i - \delta

Snell's Law:

sin⁡i=μsin⁡r=μsin⁡(i−δ)\sin i = \mu \sin r = \mu \sin (i - \delta)

Rearranging, we have:

1μ=sin⁡(i−δ)sin⁡i\frac{1}{\mu} = \frac{\sin (i-\delta)}{\sin i}

Expressing sin⁡(i−δ)\sin(i-\delta): Using the sine subtraction formula:

sin⁡(i−δ)=sin⁡icos⁡δ−cos⁡isin⁡δ\sin (i-\delta) = \sin i \cos \delta - \cos i \sin \delta

Therefore:

1μ=sin⁡icos⁡δ−cos⁡isin⁡δsin⁡i=cos⁡δ−cos⁡isin⁡isin⁡δ\frac{1}{\mu} = \frac{\sin i \cos \delta - \cos i \sin \delta}{\sin i} = \cos \delta - \frac{\cos i}{\sin i} \sin \delta

Simplify using cot⁡i\cot i: Recognize that:

cos⁡isin⁡i=cot⁡i=1tan⁡i\frac{\cos i}{\sin i} = \cot i = \frac{1}{\tan i}

Hence:

1μ=cos⁡δ−sin⁡δtan⁡i\frac{1}{\mu} = \cos \delta - \frac{\sin \delta}{\tan i}