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 min…

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asked Oct 21, 2021 28.7k views

1 Answer

Lakshya Thakur · Answer 1 · 2021-10-27T22:41:53+0000

Answer:

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