Answer:
The wavelength of the light is
m
Step-by-step explanation:
Given :
Distance between two slit
m
Distance of screen
m
Distance between central fringe to 5th bright fringe,
m
From the formula of interference,
![d \sin \theta = n \lambda](https://img.qammunity.org/2021/formulas/physics/high-school/zadlqus6da1ibkwqkgggu1ibh5wb4c9yx4.png)
Where
≅
we put
![\theta = (x)/(D)](https://img.qammunity.org/2021/formulas/physics/high-school/i4bsj62zct9hqs4s5a06w5g29y73pkihcl.png)
![(dx)/(D) = n\lambda](https://img.qammunity.org/2021/formulas/physics/high-school/z3mk7tg1xqju54dm4ertq9tvgpleb21ftk.png)
Where
5 ( given in question )
Now we have to find wavelength of the light,
![\lambda = (dx)/(nD)](https://img.qammunity.org/2021/formulas/physics/high-school/ovhddenp7yno0xnejcpd5hnvygk4ow8g31.png)
![\lambda = (0.040 * 10^(-3) * 10 * 10^(-2) )/(5 * 2)](https://img.qammunity.org/2021/formulas/physics/high-school/s6xnx9qyrjtz9xw1i0v2opm580ylzt1py3.png)
m
Therefore, the wavelength of the light is
m