Question

What is the critical angle if angle of incidence is 45 degree and angle of refraction is 25.37 degree

Answer 1

It is given that the critical angle of glass in air is 41.8o41.8o and that the refractive index of water is 1.33.1.33.

1μ=sinisinr,1μ=sin⁡isin⁡r, where,

μμ is the refractive index of glass (the ray is travelling from glass to air),

ii is the angle of incidence, and

r is the angle of refraction.

⇒1μ=sin41.8osin90o.⇒1μ=sin⁡41.8osin⁡90o.

⇒μ=1sin41.8o.⇒μ=1sin⁡41.8o.

When this glass is kept in water, its refractive index would become 1sin41.8o1.33=11.33sin41.8o=μ∗.1sin⁡41.8o1.33=11.33sin⁡41.8o=μ∗.

Then, for a ray incident (travelling from glass to water) on the glass / water interface at the critical angle,

1μ∗=sinisinr=sinisin90o=sini.1μ∗=sin⁡isin⁡r=sin⁡isin⁡90o=sin⁡i.

⇒sini=1.33sin41.8o=0.8865.⇒sin⁡i=1.33sin⁡41.8o=0.8865.

⇒i=arcsin(0.8865)=62.4352o.⇒i=arcsin⁡(0.8865)=62.4352o.


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