Light falls from glass $(μ = 1.50)$ to air. Find the angle of incidence for which the angle of deviation is $90^{\circ}$ .

Open in App

Solution

Since, $μ = 1.5$ ,

Critical Angle $= s i n^{- 1} (\frac{1}{μ})$

$= s i n^{- 1} (\frac{1}{μ}) = 41.80$

We know, the maximum attainable deviation in refraction is $(90^{\circ} - {41.8}^{\circ})$

$= {47.2}^{\circ}$

So, in this case, total internal reflection must have taken place.

In reflection,

Deviation $= 180^{\circ} - 2 i$

$= 90^{\circ}$

$\Rightarrow 2 i = 90^{\circ} \Rightarrow i = 45^{\circ}$

Suggest Corrections

Similar questions

Q. Light is incident from glass (μ = 1.50) to water (μ = 1.33). Find the range of the angle of deviation for which there are two angles of incidence.

Light is incident from glass $(μ = 1.5)$ to air. Sketch the variation of the angle of deviation $δ$ with the angle of incident i for $0 < i < 90^{\circ}$ .

Join BYJU'S Learning Program

Light falls from glass (μ=1.50) to air. Find the angle of incidence for which the angle of deviation is 90∘.

Since, μ=1.5, Critical Angle =sin−1(1μ) =sin−1 (1μ)=41.80 We know, the maximum attainable deviation in refraction is (90∘−41.8∘) =47.2∘ So, in this case, total internal reflection must have taken place. In reflection, Deviation =180∘−2i =90∘ ⇒2i=90∘⇒i=45∘

Light falls from glass $(μ = 1.50)$ to air. Find the angle of incidence for which the angle of deviation is $90^{\circ}$ .