Question

Two slits are separated by a distance of $0.5\,{\rm{mm}}$ and illuminated with light of $\lambda = 6000\,\mathop A\limits^o $. If the screen is placed $2.5\,{\rm{m}}$from the slits, the distance of the third bright image from the center will be:
(A) $1.5\;{\rm{mm}}$
(B) $3\;{\rm{mm}}$
(C) $6\;{\rm{mm}}$
(D) $9\;{\rm{mm}}$

Answer 1

Hint: First, convert all the units into standard form. Then, apply the concept fringes. Afterward, use the formula of the distance of nth bright fringes from the center to get the required result.

Complete step by step solution:
Given,
The wavelength of light used is $\lambda = 6000\,\mathop A\limits^{\rm{o}} $.
The number of fringes is $n = 3$.
The separation between the two slits is $d = 0.5\,mm$
Now, we need to convert wavelengths from angstrom to meter. That is,
$\begin{array}{c}\lambda = 6000\,\mathop A\limits^o \\ = 6000 \times {10^{ - 10}}\,m\end{array}$

The separation between the slits needs to be converted from millimeter to meter. That is,
$\begin{array}{c}d = 0.5\,mm\\ = 5 \times {10^{ - 4}}\,m\end{array}$
The distance between slits and screen is,
$D = 2.5\,m$

The expression of the distance of nth bright fringes from the center is given by,
${y_n} = \dfrac{{n\lambda D}}{d}$ …… (1)

Here, $n,\lambda ,D$and $d$ are the number of fringes, wavelength, the distance between the screen and the slits and separation between the slits.
Substitute the value of $n,\lambda ,D$and $d$ in equation (1),
$\begin{array}{l}{y_n} = \dfrac{{3 \times 6000 \times {{10}^{ - 10}} \times 2.5}}{{5 \times {{10}^{ - 4}}}}\\{y_n} = 9\,mm\end{array}$
Thus, the distance of the third bright image from the center is ${y_n} = 9\,mm$.
So, the correct option is (D).

Note:
Students should take units of each quantity. All the students should take care of terms of denominator and numerator in the formula of the distance of nth bright fringes.