A thin film of refractive index 1.5 and thickness `4 xx 10^(-5)` cm is illuminated by light normal to the surface. What wavelength within the visible spectrum will be intensified in the reflected beam?

Options

A. (a) `4800Å`
B. (b) `5800Å`
C. (c) `6000Å`
D. (d) `6800Å`

Answer

A

Explanation

Condition for observing bright fringe is:
`2^(nd)=(m+1/2)lambda`
`:. lambda=(2nd)/(m+1/2)=(2xx1.5xx4xx10^-5)/(m+1/2)=(12xx10^-5)/(m+1/2)`
The integer m that gives the wavelength in the visible region (`4000Å` to `Å`) is `m=2`. In that case, `lambda=(12xx10^-5)/(2+1/2)=4.8xx10^-5=4800Å`.