Question

You have just installed a new bathroom in your home. Your shower doors have frosted glass to provide privacy for the person using the shower. The frosted surface is on the outside of the shower door, facing the rest of the bathroom. The frosting is done by acid etching the surface so that light incident on the rough surface is scattered in all directions. Proud of your new bathroom, you take a photo of it with your smartphone. You notice in the photograph that you can see a reflection of the flash in the shower doors and the reflection is surrounded by a halo of light. Curious, you turn on a laser pointer and aim it at the shower door. Looking closely at the reflection, you again see a halo that consists of a dark area surrounding the reflection of the pointer and then an area of brightness outside this dark ring. You grab a micrometer and a ruler and measure the thickness of the glass to be 7.00 mm and the inner radius of the bright halo to be 10.5 mm. From these measurements, you determine the index of refraction of the glass.

I am coming out for the index of refraction for the glass at .6668. Using Snell's Law and a triangle from center to back pane of glass for refraction angle. use that angle and 90-refraction angle for my incident of ray angle. n1sin(X)=n2sin(Y) where n1=index refraction of air, sin(x)=incident ray angle, n2= what i am solving for, sin(Y)= refraction angle. thx for any ideas.

Answer 1

I'm just giving you a brief outline of the solution.

The is due to interference of lights reflected from the outer surface of the glass and inner surface.

Now please take a look at the figures below,

fig 2 illustrates the process.

Considering the whole phenomenon as interference due to a thin film, the formula for the maxima of the interference pattern is there, just put the numbers there and get your required value of refraction angle.

in the formula however, 'd' stands for the thickness of the glass, 'n' is the refraction index of the glass, m=0 sets the 1st maxima. Just put the wavelength of the light of the laser you are using in the equation and get the answer.