Question

The Light of wavelength $600\;{\text{nm}}$ is incident normally on a slit of width $3\;{\text{mm}}$. Calculate angular width of the central maximum on a screen kept $3\;{\text{m}}$ away from the slit.

Answer 1

Hint:The wavelength and the normal width of slit are given in the question. The angular width is calculated by taking the ratio of linear width to perpendicular distance of source from the slits.

Complete step by step answer:

In this question we have given the following values that is the wavelength of the light is $600\;{\text{nm}}$, the width of the silt is $3\;{\text{mm}}$ and the screen is placed $3\;{\text{m}}$ away from the slit. We need to calculate the angular width of the central maximum.

We know that the formula for linear width can be expressed as,
${\text{linear}}\;{\text{width}} = \dfrac{{2\lambda D}}{a}$
Here, $\lambda $ is the wavelength of the incident light wave, $D$ is the perpendicular distance of the slits from the source and $a$ is slit width.
As we know that the angular width of central maxima can be calculated by ratio of linear width to the given slit width.

Now, we write the expression for the angular width of central maxima.
\[{\text{Angular width}} = \dfrac{{{\text{linear Width}}}}{D}\]

Now we substitute the value of linear width in above equation to calculate the value of angular width of central maxima as

\[ \Rightarrow {\text{Angular width}} = \dfrac{{\dfrac{{2\lambda D}}{a}}}{D}\]

Further simplify the above equation we get,


\[ \Rightarrow {\text{Angular width}} = \dfrac{{2\lambda }}{a}\]
Now, we substitute the value of $\lambda $ and $a$ in above equation.
\[ \Rightarrow {\text{Angular width}} = \dfrac{{2\left( {600\;{\text{nm}}} \right)}}{{3\;{\text{mm}}}}\]
Now, we convert unit of nanometer to millimeter as
\[ \Rightarrow {\text{Angular width}} = \dfrac{{\left( {1200\;{\text{nm}}} \right)\left( {\dfrac{{{{10}^{ - 6}}{\text{mm}}}}{{1\;{\text{nm}}}}} \right)}}{{\left( {3\;{\text{mm}}} \right)}}\]
After simplification we get,
\[\therefore {\text{Angular width}} = 4 \times {10^{ - 4}}\;{\text{rad}}\]

Therefore, the angular width of central maxima is \[4 \times {10^{ - 4}}\;{\text{rad}}\].


Note:Remember that we have to calculate the value of angular width of central maxima. In the question an extra data is given which is not necessary to be used. So neglect the distance of source to slit as the value of this distance is cancelled out in the formula of angular width.

One more thing which should be avoided in this calculation is that it doesn’t put extra effort in calculating the linear width and then calculating the angular width. Just substitute the values in the formula of angular width itself.