A ray of light passes from a denser medium to a rarer medium at an angle of incidence i. The reflected and refracted rays make an angle of 90° with each other. The angle of reflection and refraction are respectively r and r prime. The critical angle is given by:

Question

A ray of light passes from a denser medium to a rarer medium at an angle of incidence i. The reflected and refracted rays make an angle of 90° with each other. The angle of reflection and refraction are respectively r and r prime. The critical angle is given by: (A)  sin -1( cot r) (B)  tan-1( sin i) (C)  sin -1( tan r prime) (D)  sin -1( tan r)

Tardigrade Admin · Accepted Answer

r+r prime+90°=180° ⇒ r prime=90-r=90-i n1 sin i=n2 sin r prime=n2 sin (90-i) n1 sin i=n2 cos i ⇒ tan i=(n2/n1) Now sin C=(n2/n1)= tan i ⇒ C= sin -1( tan i)= sin -1( tan r)

Q. A ray of light passes from a denser medium to a rarer medium at an angle of incidence $i$ . The reflected and refracted rays make an angle of $9 0^{\circ}$ with each other. The angle of reflection and refraction are respectively $r$ and $r^{'}$ . The critical angle is given by:

$sin^{- 1} (cot r)$

$tan^{- 1} (sin i)$

$sin^{- 1} (tan r^{'})$

$sin^{- 1} (tan r)$

Q. A ray of light passes from a denser medium to a rarer medium at an angle of incidence i. The reflected and refracted rays make an angle of 90∘ with each other. The angle of reflection and refraction are respectively r and r′. The critical angle is given by:

sin−1(cotr)

tan−1(sini)

sin−1(tanr′)

sin−1(tanr)

r+r′+90∘=180∘ ⇒r′=90−r=90−i n1​sini=n2​sinr′=n2​sin(90−i) n1​sini=n2​cosi ⇒tani=n1​n2​​ Now sinC=n1​n2​​=tani ⇒C=sin−1(tani)=sin−1(tanr)

Q. A ray of light passes from a denser medium to a rarer medium at an angle of incidence $i$ . The reflected and refracted rays make an angle of $9 0^{\circ}$ with each other. The angle of reflection and refraction are respectively $r$ and $r^{'}$ . The critical angle is given by:

$sin^{- 1} (cot r)$

$tan^{- 1} (sin i)$

$sin^{- 1} (tan r^{'})$

$sin^{- 1} (tan r)$