Question

In a double-slit experiment, the optical path difference between the rays from two coherent sources at a point P on side of the central bright band is $7.5 \times {10^{ - 6}}m$ and at a point Q on the other side of the central bright band is $1.8 \times {10^{ - 6}}m$. How many bright and dark bands are observed between the points P and Q if the wavelength of light used is $6 \times {10^{ - 7}}m$?

Answer 1

Hint: Here, we will use the Young’s double slit formula to find the number of dark and bright bands. Also, with the help of the law of power and exponents and the fractions will simplify the values for the resultant answer.

Complete step by step answer:
Let us consider the coherent sources at a point P are \[\Delta {x_1} = 7.5 \times {10^{ - 6}}m\]
And the coherent sources at a point P be \[\Delta {x_2} = 1.8 \times {10^{ - 6}}m\]
Also, the wavelength of the light is $\lambda = 6 \times {10^{ - 7}}m$
The above can be re-written as $\lambda = 0.6 \times {10^{ - 6}}m$ (To make all the given terms in the form of same multiple)
Now, by using the Young’s double slit formula –
$\Delta {x_1} = (2{n_1} - 1)\dfrac{\lambda }{2}$
Place the values in the above equation –
$ 7.5 \times {10^{ - 6}} = (2{n_1} - 1)\left( {\dfrac{{0.6 \times {{10}^{ - 6}}}}{2}} \right)$
Bring all the terms at one side- when the term is in the multiplicative form at one side is moved to other side it goes to the division and vice-versa.
$ \left( {7.5 \times {{10}^{ - 6}}} \right) \times \left( {\dfrac{2}{{0.6 \times {{10}^{ - 6}}}}} \right) = (2{n_1} - 1)$
Same terms from the numerator and the denominator cancel each other-
$ \left( {\dfrac{{2 \times 7.5}}{{0.6}}} \right) = (2{n_1} - 1)$
Simplify the above equation –
$ \Rightarrow 25 = (2{n_1} - 1)$
When the term is moved from one side to another, sign of the term is also changed. Positive term changes to negative and vice-versa.
\[
25 + 1 = 2{n_1} \\
\Rightarrow 26 = 2{n_1} \\
\Rightarrow {n_1} = \dfrac{{26}}{2} \\
\Rightarrow {n_1} = 13 \\
 \]
Similarly, $\Delta {x_2} = (2{n_2} - 1)\dfrac{\lambda }{2}$
Place the values in the above equation –
$ \Rightarrow 1.8 \times {10^{ - 6}} = (2{n_2} - 1)\left( {\dfrac{{0.6 \times {{10}^{ - 6}}}}{2}} \right)$
Bring all the terms at one side- when the term is in the multiplicative form at one side is moved to other side it goes to the division and vice-versa.
$ \left( {1.8 \times {{10}^{ - 6}}} \right) \times \left( {\dfrac{2}{{0.6 \times {{10}^{ - 6}}}}} \right) = (2{n_2} - 1)$
Same terms from the numerator and the denominator cancel each other-
$ \left( {\dfrac{{2 \times 1.8}}{{0.6}}} \right) = (2{n_2} - 1)$
Simplify the above equation –
$
\dfrac{{3.6}}{{0.6}} = (2{n_2} - 1) \\
\Rightarrow 6 = (2{n_2} - 1) \\
 $
When the term is moved from one side to another, the sign of the term is also changed. Positive term changes to negative and vice-versa.
\[
 6 + 1 = 2{n_2} \\
\Rightarrow 7 = 2{n_2} \\
\Rightarrow {n_2} = \dfrac{7}{2} \\
\Rightarrow {n_2} = 3.5 \\
\Rightarrow {n_2} \approx 3 \\
 \]

Hence, we have total bright and dark bands $13 + 3 = 16$ bands in the above double slit experiment.

Note:
Know the difference between the coherent and the incoherent sources. Coherent sources are the sources in which waves are in the phase or have the definite phase relationships whereas the incoherent sources have random phases. Also, know the properties of the power and exponent and its law for the application.