Question

Light with a wavelength of 600 nm shines onto a single slit, and the diffraction pattern is observed on a screen 2.5 m away from the slit. The distance, on the screen, between the dark spots to either side of the central maximum in the pattern is 25 mm. (a) What is the distance between the same dark spots when the screen is moved so it is only 1.5 m from the slit

Answer 1

Answer:

"6.67 mm" is the right solution.

Explanation:

The given values are:

• L = 2.5 m
• y = .0125
• λ = 600 nm

As we know, the equation

⇒  $\frac{y}{L} =\frac{x \lambda}{a}$

On substituting the values, we get

⇒  $\frac{.0125}{2.5}=\frac{(1)(600\times 10^{-9}) }{a}$

On applying cross multiplication, we get

⇒  $.0125a=2.5 (600\times 10^{-9})$

⇒          $a=\frac{2.5(600\times 10^{-9})}{.0125}$

⇒             $=1.2\times 10^{-4} \ m$

For new distance, we have to put this value of "a" in the above equation,

⇒  $\frac{y}{1.5} =\frac{(1)(600\times 10^{-9})}{1.2\times 10^{-4}}$

⇒  $(1.2\times 10^{-4})y=1.5(600\times 10^{-9})$

⇒                     $y=\frac{1.5(600\times 10^{-9})}{1.2\times 10^{-4}}$

⇒                        $=3.22\times 10^{-3} \ m$

The total distance will be twice the value of "y", we get

=  $6.67\times 10^{-3} \ m$

or,

=  $6.67 \ mm$