question on critical angle...

inverse tan

hi,
i came across a question on critical angle.the question is:
light travelling in air strikes a glass plate at an angle of 33 deg. measured between the incoming ray and the glass, part of the beam is reflected and part is refracted. if the refracted and reflected rays make an angle of 90 deg. with each other,what is the critical angle for this glass?



is total internal reflection possible when light travels from a rarer medium to a denser medium?what does critical angle mean here?

-karthy

critical angle is

sininverse(tan(33degrees))=40.497degrees

When light travels from an "optically lighter" medium to an "optically denser" medium total internal reflection "CANNOT" take placel.....IN YOUR CASE THE QUESTION OF CRITICAL ANGLE IS FOR THE CASE WHEN LIGHT TRAVELS FROM GLASS TO AIR AS ONLY THEN CRITICAL REFLECTION TAKES PLACE FOR THE CRITICAL ANGLE TO HAVE A MEANING AT ALL!!

The answer to the first question corresponds to v_naren's solution but the following solution also accounts for potential differences in the index of refraction for different glass types (an index of refraction of 1.54 for glass is common, but not all glass has an index of refraction of 1.54). Since the reflected angle was given, and the index of refraction can be extracted, it is prudent to explore that possibility.

The critical angle can be calculated by taking the inverse-sine of the ratio of the indices of refraction of the reflected and incident mediums.

Since the incident angle is 33 degrees the reflected angle is therefore 33 degrees. The problem states that the reflected and refracted angles are separated by 90 degrees. That leaves a reflected angle of 57 degrees with respect to the plane of the glass.

However, for the law of refraction shown below, where Ni here refers to the index of refraction for the incident and Nr for the reflected material, the angles are relative to the normal (perpendicular) to the glass.

Therefore Ni = 57 and Nr = 33.

Ni * sin(i) = Nr * sin(r)

Air has a Ni of 1.0.

Therefore Nr= (Ni*sin(i)) / sin(r) = 1*sin(57) / sin(33) = 1.5399 = 1.54

Then the ratio of Nr to Ni = 1.54. The inverse sin of 1.54 (and the critical angle) is 40.496 degrees or 40.5 degrees.

The second question is vague. What does "optically dense" mean? Only if it correlates to index of refraction can the question be answered. But if so, as v_naren stated the critical angle only exists for a transition from a a higher to lower index of refraction. Otherwise in the formula above you would be taking the arcsine of a number greater than 1 -- which is a complex number.