Question

The distance between two consecutive bright bands in a bi-prism experiment is $0.32\,{\text{mm}}$ when red light of wavelength $6400\,\mathop {\text{A}}\limits^{\text{o}}$ is used. By how much will this distance change if the light is substituted by the blue light of wavelength $4800\,\mathop {\text{A}}\limits^{\text{o}}$ with the same setting?

Answer 1

Use the formula for the fringe width in a bi-prism experiment. This formula gives the relation between the fringe width, distance between the slit and eyepiece and distance between the two virtual sources formed by the bi-prism. Calculate the fringe width using this formula and then calculate the change in fringe width between the two values of fringe width.

Formula used:
The fringe width $\beta$ in a bi-prism experiment is given by
$\beta = \dfrac{{\lambda D}}{d}$ …… (1)
Here, $\lambda$ is the wavelength of the light used, $D$ is the distance between the slit and eyepiece and $d$ is the distance between the two virtual sources formed by the bi-prism.

Complete step by step answer:
We have given that the distance between the two consecutive bright bands of fringe width in the bi-prism experiment is $0.32\,{\text{mm}}$ when the red light is used.
${\beta _R} = 0.32\,{\text{mm}}$
The wavelength of the red light used is $6400\,\mathop {\text{A}}\limits^{\text{o}}$ and the wavelength of the blue light is $4800\,\mathop {\text{A}}\limits^{\text{o}}$.
${\lambda _R} = 6400\,\mathop {\text{A}}\limits^{\text{o}}$
${\lambda _B} = 4800\,\mathop {\text{A}}\limits^{\text{o}}$
Rewrite equation (1) for the fringe width when the red light is used.
${\beta _R} = \dfrac{{{\lambda _R}D}}{d}$ …… (2)
Rewrite equation (1) for the fringe width when the blue light is used.
${\beta _B} = \dfrac{{{\lambda _B}D}}{d}$ …… (3)

Divide equation (3) by equation (2).
$\dfrac{{{\beta _B}}}{{{\beta _R}}} = \dfrac{{\dfrac{{{\lambda _B}D}}{d}}}{{\dfrac{{{\lambda _R}D}}{d}}}$
$\Rightarrow \dfrac{{{\beta _B}}}{{{\beta _R}}} = \dfrac{{{\lambda _B}}}{{{\lambda _R}}}$
$\Rightarrow {\beta _B} = \dfrac{{{\lambda _B}}}{{{\lambda _R}}}{\beta _R}$
Substitute $4800\,\mathop {\text{A}}\limits^{\text{o}}$ for ${\lambda _B}$ , $6400\,\mathop {\text{A}}\limits^{\text{o}}$ for ${\lambda _R}$ and $0.32\,{\text{mm}}$ for ${\beta _R}$ in the above equation.
$\Rightarrow {\beta _B} = \dfrac{{4800\,\mathop {\text{A}}\limits^{\text{o}} }}{{6400\,\mathop {\text{A}}\limits^{\text{o}} }}\left( {0.32\,{\text{mm}}} \right)$
$\Rightarrow {\beta _B} = 0.24\,{\text{mm}}$
Hence, the fringe width when the blue light is used is $0.24\,{\text{mm}}$.

Let us now calculate the change in the fringe width $\Delta \beta$ when blue light is used.
$\Delta \beta = {\beta _R} - {\beta _B}$
Substitute $0.32\,{\text{mm}}$ for ${\beta _R}$ and $0.24\,{\text{mm}}$ for ${\beta _B}$ in the above equation.
$\Delta \beta = \left( {0.32\,{\text{mm}}} \right) - \left( {0.24\,{\text{mm}}} \right)$
$\therefore \Delta \beta = 0.08\,{\text{mm}}$

Hence, the change in the distance between the two consecutive bright bands when blue light is used is $0.08\,{\text{mm}}$.

Note: The students may think that why the values of $D$ and $d$ are taken the same in the expression for the fringe width for both the red and blue light. But the students should keep in mind that the values of $D$ and $d$ remain the same for both the lights used as the setting of the experiment is the same for both the lights. The students should keep in mind that we have to be asked to calculate the change in fringe width and not the fringe width for the blue light.