Question

White light may be considered to be a mixture of waves of $\lambda$ ranging between $3900\mathring{A}\text{and}7800\mathring{A}.$ An oil film of thickness $10000\mathring{A}$ is examined normaly by the reflected light. If $\mu =1.4,$ then the film appears bright for

• A. $4000\mathring{A},4667\mathring{A},5600\mathring{A},7000\mathring{A}$
• B. $4308\mathring{A},5091\mathring{A},6222\mathring{A}$
• C. $4000\mathring{A},5091\mathring{A},5600\mathring{A}$
• D. $4667\mathring{A},6222\mathring{A},7000\mathring{A}$

Answer 1

The correct option is B $4308\mathring{A},5091\mathring{A},6222\mathring{A}$

For the given condition,
$2\mu t-\frac{\lambda }{2}=n\lambda$
$2\mu t=(n+1/2)\lambda$
$\lambda =\frac{2\mu t}{n+1/2}=\frac{2\times 1.4\times 10000}{n+1/2}=\frac{28000}{n+1/2}$
Now, for
$n=0,\lambda =56000\mathring{A}$
$n=1,\lambda =15333\mathring{A}$
$n=2,\lambda =11200\mathring{A}$
$n=3,\lambda =8000\mathring{A}$
$n=4,\lambda =6222\mathring{A}$
$n=5\lambda =5091\mathring{A}$
$n=6,\lambda =4308\mathring{A}$
Thus, we can see here the wavelengths lie between the range $3900\mathring{A}$ and $7800\mathring{A}$ are
$\lambda =4308\mathring{A},5091\mathring{A},6222\mathring{A}$
Hence, option (B) is right.