Question

One light wave is incident upon a

plate of refracting index m . Incident

angle i , for which refractive &

reflective waves are mutually

perpendicular will be a. i = 450b. i=sin

−1(?)c. i = coesc−1(?)d. ?=tan−1

(?)​

Answer 1

I think option d)tan^-1

i + 90°+ r =180°

i+ r =90°

r=90° -i

(miu) u=sin i /sin r

=sin i/ sin(90°- i)

=sin i/ cos i

(miu) u=tan i

i=tan i ^-1(u)

Answer 2

Given:

Refractive index of the medium = μ

Angle made by reflected and refracted ray = 90°

To Find: Angle of Incidence

Calculation:

From the figure below:

• PQ is the interface between two medium.
• MN is the normal at the interface.
• AO is the incident ray that makes angle i with the normal, which means i is the angle of incidence.
• OB is the reflected ray which makes angle of reflection with the normal which is equal to i.
• OC is the refracted ray which makes angle r with the normal.

For straight line MN:

∠i + 90° + ∠r = 180°

∠i + ∠r = 90°

∠r = 90° - ∠i

sin r = sin (90 - i)

sin r = cos i

Now, refractive index is given as:

μ = $\frac{sin i}{sin r}$

μ = $\frac{sin i}{cos i} = tani$

i = $tan^{-1}$μ

Answer:

Angle of incidence i = $tan^{-1}$μ

Your question is improper. You might be referring to the question below:

"One light wave is incident upon a plate of refracting index μ. Incident angle i, for which refractive & reflective waves are mutually  perpendicular will be

a. i = 45°

b. i = $sin^{-1}$μ

c. i = $cosec^{-1}$μ

d. i = $tan^{-1}$μ"