Physics
Physics
1st Edition
Walker
ISBN: 9780133256925
Textbook solutions

All Solutions

Page 652: Practice Problems

Exercise 20
Step 1
1 of 2
From the condition for destructive interference we have

$$
2nt=mlambda.
$$
This yields for the thickness

$$
t=frac{mlambda}{2n}.
$$
Minimum thickness is obtained when we put $m=1$ yielding

$$
t=frac{lambda}{2n} = frac{662times10^{-9}text{ m}}{2times 1.33} =249text{ nm}.
$$

Result
2 of 2
Click here for the solution.
Exercise 21
Step 1
1 of 2
From the condition for destructive interference we have

$$
2nt=mlambda.
$$
This yields for the thickness

$$
t=frac{mlambda}{2n}.
$$
Minimum thickness is obtained when we put $m=1$ yielding

$$
t=frac{lambda}{2n} = frac{670times10^{-9}text{ m}}{2times 1.30} =258text{ nm}.
$$

Result
2 of 2
Click here for the solution.
Exercise 22
Step 1
1 of 2
$textbf{Given values:}$

$$
begin{align*}
t &= 183 text{ nm} \
n &= 1.35 \
m &= 0
end{align*}
$$

The wavelength of the beam of light which experiences constructive interference on a soap film can be obtain by applying the relation for constructive interference in a thin film :

$$
begin{align*}
2 n t &= left( m + dfrac{1}{2} right) lambda \
lambda &= dfrac{2 n t}{m + dfrac{1}{2}} \
&= dfrac{2 (1.35) (183 text{ nm})}{0 + dfrac{1}{2}}
end{align*}
$$

$$
{boxed{lambda = 988.2 text{ nm}}}
$$

Result
2 of 2
$$
{lambda = 988.2 text{ nm}}
$$
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