In a Fraunhofer diffraction experiment at a single slit using a light of wavelength 400nm, the first minimum is formed at an angle of \[{30^ \circ }\]. The direction $\theta $ of the first secondary maximum is given by:
a. ${\sin ^{ - 1}}\left( {\dfrac{2}{3}} \right)$
b. ${\sin ^{ - 1}}\left( {\dfrac{3}{4}} \right)$
c. ${\sin ^{ - 1}}\left( {\dfrac{1}{4}} \right)$
d. $ta{n^{ - 1}}\left( {\dfrac{2}{3}} \right)$

Hint: From the path difference relation for minima try to find the slit width. Use that to find the direction for secondary maximum.

Formula Used:
For minima of a diffraction:
$a\sin {\theta _{\min }} = n\lambda $ -----(1)
Where,
a is width of the slit,
n is order of the minima,
${\theta _{\min }}$ is angle for ${n^{th}}$ minima formation,
$\lambda $ is the wavelength of the light.

For maxima of a diffraction:
$a\sin {\theta _{\max }} = \left( {n + \dfrac{1}{2}} \right)\lambda $ ------(2)
${\theta _{\max }}$ is angle for ${n^{th}}$maxima formation.

Complete answer:
Given:
Wavelength of the light used, $\lambda = 400nm$.
For n=1 i.e. for first minima ${\theta _{\min }} = {30^ \circ }$.

To find: direction angle for first secondary maxima.

> Step 1
Substitute the value of n, $\lambda $and ${\theta _{\min }}$in eq(1) to get a:
$
  asin\left( {{{30}^ \circ }} \right) = 1 \times 400 \\
   \Rightarrow a \times \dfrac{1}{2} = 400 \\
   \Rightarrow a = 400 \times 2 = 800nm \\
 $

> Step 2
For first secondary maxima, $n = 1$.
Now, use the value of a in eq(2) to get ${\theta _{\max }}$:
$
  800 \times sin{\theta _{\max }} = \left( {1 + \dfrac{1}{2}} \right) \times 400 \\
   \Rightarrow sin{\theta _{\max }} = \dfrac{3}{2} \times \dfrac{{400}}{{800}} \\
   \Rightarrow {\theta _{\max }} = {\sin ^{ - 1}}\left( {\dfrac{3}{4}} \right) \\
 $

Hence, the direction $\theta $ of the first secondary maximum is given by ${\sin ^{ - 1}}\left( {\dfrac{3}{4}} \right)$.

Note: This problem can be solved without using the value of $\lambda$ and without finding the value of a. Just take the ratio of eq(1) and eq(2) after inserting proper values of n and ${\theta _{\min }}$. You’ll see:
$
  \dfrac{{a\sin {\theta _{\max }}}}{{a\sin ({{30}^ \circ })}} = \dfrac{{\left( {1 + \dfrac{1}{2}} \right)\lambda }}{{1 \times \lambda }} \\
   \Rightarrow \sin {\theta _{\max }} = \dfrac{3}{2} \times \dfrac{1}{2} \\
   \Rightarrow {\theta _{\max }} = {\sin ^{ - 1}}\left( {\dfrac{3}{4}} \right) \\
 $
And this is the same answer you got previously but this time with a lot less amount of calculations.

Also, many students get confused with the term ‘first secondary maxima’. Always remember that the first secondary maxima is the first maxima beside the central one, not the central one itself. So, for this maxima the value of n will be 1 not 0.