KCET · Physics · Wave Optics
In Young's double slit experiment, an electron beam is used to produce interference fringes of width \(\beta_1\). Now the electron beam is replaced by a beam of protons with the same experimental set-up and same speed. The fringe width obtained is \(\beta_2\). The correct relation between \(\beta_1\) and \(\beta_2\) is
- A \(\beta_1=\beta_2\)
- B No fringes are formed
- C \(\beta_1 \lt \beta_2\)
- D \(\beta_1\gt\beta_2\)
Answer & Solution
Correct Answer
(D) \(\beta_1\gt\beta_2\)
Step-by-step Solution
Detailed explanation
In Young's double slit experiment triangle
\(\beta=\frac{D \lambda}{d}\)
\(\Rightarrow \quad \frac{\beta_2}{\beta_1}=\frac{\lambda_2}{\lambda_1}\) ....(i)
\(\Rightarrow \therefore \quad \frac{\lambda_2}{\lambda_1} \lt 1\) ...(ii)
From Eqs. (i) and (ii), we get
\(\frac{\beta_2}{\beta_1} \lt 1 \Rightarrow \beta_2 \lt \beta_1\)
\(\beta=\frac{D \lambda}{d}\)
\(\Rightarrow \quad \frac{\beta_2}{\beta_1}=\frac{\lambda_2}{\lambda_1}\) ....(i)
\(\Rightarrow \therefore \quad \frac{\lambda_2}{\lambda_1} \lt 1\) ...(ii)
From Eqs. (i) and (ii), we get
\(\frac{\beta_2}{\beta_1} \lt 1 \Rightarrow \beta_2 \lt \beta_1\)
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