Question

The distance between the slit and the bi-prism and that of between bi-prism and screen each is 0.4 m. The obtuse angle of bi-prism is ${{179}^{o}}$ and the refractive index is 1.5. If the fringe width is $1.8\times {{10}^{-4}}m$ then the wavelength of light will be:
$\begin{align}
  & \text{A}\text{. }7850\overset{o}{\mathop{A}}\, \\
 & \text{B}\text{. }6930\overset{o}{\mathop{A}}\,\text{ } \\
 & \text{C}\text{. }5890\overset{o}{\mathop{A}}\, \\
 & \text{D}\text{. }3750\overset{o}{\mathop{A}}\, \\
\end{align}$

Answer 1

Hint: We are given the distance from slit to bi – prism and from bi – prism to screen. We are also given the obtuse angle and refractive index of the prism and fringe width. To find the wavelength of the light we have an equation. By calculating all the variables in the equation using the given values and substituting them in the equation, we get the solution.

Formula used:
Wavelength, $\lambda =\dfrac{d\beta }{D}$

Complete answer:
In this question there is a bi – prism.
The distance between the slit and the bi – prism is given as 0.4 m and the distance from the bi – prism to the screen is also 0.4 m.
Let ‘$\beta $’ be the fringe width. It is given as,
$\beta =1.8\times {{10}^{-4}}m$
The obtuse angle of the bi – prism is given as ${{179}^{0}}$
Let ‘A’ be the acute, then each acute angle of the prism will be,
$A=\dfrac{{{180}^{o}}-{{179}^{o}}}{2}={{0.5}^{o}}$
By converting this into radians, we get
$A=0.5\times \dfrac{\pi }{180}=\dfrac{\pi }{360}rad$
We know that deviation of the light from source ‘$\delta $ ’ is given as,
$\delta =A\left( \mu -1 \right)$, were ‘$\mu $’ is the refractive index of the prism given as 1.5
By substituting the values we get,
$\Rightarrow \delta =\dfrac{\pi }{360}\times \left( 1.5-1 \right)$
$\Rightarrow \delta =0.004363m$
Then the deviation of each beam of light is given as the product of ‘$\delta $’ and distance between source and prism.
$\omega =\delta \times 0.4$
$\Rightarrow \omega =A\left( \mu -1 \right)0.4$
$\Rightarrow \omega =0.004363\times 0.4$
$\Rightarrow \omega =0.001745m$
Now we have the deviation of each beam of light. We know that the distance between imaginary sources is twice the deviation of each beam of light, i.e.
$d=2\times \omega $
$\Rightarrow d=2\times 0.001745$
$\Rightarrow d=0.0035m=3.5mm$
Let ‘D’ be the total distance from source to screen. Therefore,
$D=0.8m$
We are asked to find the wavelength of the light.
The equation for wavelength is given by,
$\lambda =\dfrac{d\beta }{D}$, where ‘$\lambda $’ is wavelength and ‘$\beta $’ is the fringe width.
Now let us substitute the known values in the above equation, thus we get
$\Rightarrow \lambda =\dfrac{0.0035\times 1.8\times {{10}^{-4}}}{0.8}$
By simplifying this we get,
$\lambda \approx 7850\overset{o}{\mathop{A}}\,$

So, the correct answer is “Option A”.

Note:
A bi – prism is simply a triangular prism in which the refracting angles will be approximately ${{1}^{o}}$. It basically consists of two prisms whose bases are joined together to form an isosceles triangle.
A bi – prism is usually used to determine interference fringes and to find wavelength of a monochromatic light source. It is also used to find the thickness of a glass sheet kept on a source of light.