Question

The critical angle of light in a certain substance 45.what the polarising angle

Answer 1

Light doesn't have a critical angle "in" a certain substance. It has a critical angle of entry into (or exit from) a certain substance, to or from a different substance. But never mind: your instructor's native language is not English, right?

Anyway, the critical angle C is given by
sin(C) = n2/n1, where the n's are
the refractive indices of the two substances,
and the "n1" is definitely the
larger index of refraction.

The polarization angle P is given by
tan(P) = n2/n1, but here the "n1" is the index
for the initial medium, which may well be
the medium with the smaller index of refraction.

That's why the problem statement is so poor! - "in" a certain medium is nonsense. Relying on the likelihood that the desired polarization angle has the light passing from the lower-n medium into the higher-n medium, I'm going to rewrite the critical angle as arcsin(n1/n2), so that n2 now has the high n.

If the critical angle is 45 degrees, then n1/n2 is 0.7071, and n2/n1 is sqrt(2), so that
tan(P) = sqrt(2) and P = 54.7 degrees.

Mark it as brainliest plz..