It is desired to lessen the reflection of light by the outer surfaces of the objective lenses of a pair of binoculars. These surfaces are the foremost surfaces that light hits on entering the binoculars. The lenses are prepared of crown glass. To lessen the reflection the lenses should be coated with a transparent medium of the right thickness and of the right index of refraction. For utmost destructive interference of the reflected light the value of the index of refraction of the coating.

Answer:

Acceptable we want to put a quarter-wave thickness of optical coating on the lens to cause a path difference of half a wavelength consequently that we get destructive interference between the light path 1 light that reflects off the first interface (the air/coating interface) and the light path 2 light that reflects off the second interface (the coating/glass interface).

What we typically don't talk about in discussing thin-film interference is the amplitudes. However to get the most destructive interference where they do interfere we want the amplitudes to be the same. Then a crest of path 1 light is entirely cancelled by a trough of path 2 light. Now the bigger the dissimilarity between the indices of refraction of the two media on either side of an interface the greater the amount of reflection from that interface.

Presume we had the index of refraction of the coating exactly midway between that of air and that of the crown class. Additional suppose that 10% of the light is reflected off each surface. The value isn't important but the fact that both values are the same is reasonable since the difference ncoating-nair is under the given circumstances the same as the difference nglassncoating. If 10% is replicated off the first surface only 90% gets through. After that 10% of that 90% which is only 9% of the original intensity is reflected off the second surface and only 90% of that or 8.1% of the original intensity makes it back out through the first surface. Therefore where the light from the two paths joins together again back in the air after reflection we have 10% of the original intensity interfering with 8.1% of the original intensity consequently the cancellation will not be perfect (even if the coating is of the perfect thickness). If we reduce the index of refraction of the coating a little bit making it a little closer to the index of refraction of air we decrease the reflection at the first surface as well as increase the amount that gets through that surface and the percentage of that amount that is reflected off the second surface and then makes it back through the first surface. With the correct choice of the coating index of refraction extrapolating upon the given example we can bring that 10% down and the 8.1% up just enough to have them meet in the middle. If they are both equivalent we are able to get perfect destructive interference.

Therefore to maximize the destructive interference of the reflected light we need the index of refraction of the coating to be in between the nair and nglass, and we need it to be a little bit closer in value to nair than it is to nglass.