Question

A screen is placed 1.20m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40cm wide-that is, the two first-order diffraction minima are separated by 1.40cm

What is the distance between the two second-order minima?

Answer 1

2.8 cm

Step-by-step explanation:

`y_1` = Separation between two first order diffraction minima = 1.4 cm

D = Distance of screen = 1.2 m

m = Order

Fringe width is given by

`\beta_1=(y_1)/(2)\\\Rightarrow \beta_1=(1.4)/(2)\\\Rightarrow \beta_1=0.7\ cm`

Fringe width is also given by

`\beta_1=(m_1\lambda D)/(d)\\\Rightarrow d=(m_1\lambda D)/(\beta_1)`

For second order

`\beta_2=(m_2\lambda D)/(d)\\\Rightarrow \beta_2=(m_2\lambda D)/((m_1\lambda D)/(\beta_1))\\\Rightarrow \beta_2=(m_2)/(m_1)\beta_1`

Distance between two second order minima is given by

`y_2=2\beta_2`

`\\\Rightarrow y_2=2(m_2)/(m_1)\beta_1\\\Rightarrow y_2=2(2)/(1)* 0.7\\\Rightarrow y_2=2.8\ cm`

The distance between the two second order minima is 2.8 cm