Q.45. In a Young double slit experiment, the wavelength of incident of incident light is , the separ

Q: What is the correct answer for this exam Physics Q45?

The correct answer is C. Given wavelength $\lambda = 6000 \mathring{A}$. Slit separation $d = 5 cm = 5 \times 10^{-2} m$. Distance to screen $D = 50 cm = 0.5 m$. The resultant intensity at point P is equal to the intensity due to individual slits. Let $I_0$ be the intensity due to a single slit. Then the resultant intensity $I = 4 I_0 \cos^2(\phi/2)$, where $\phi$ is the phase difference. Given $I = I_0$. So $I_0 = 4 I_0

Q: Q.45. In a Young double slit experiment, the wavelength of incident of incident light is , the separation between slits and is 5 cm the distance between slits plane and screen is 50 cm, as shown in…

The correct answer is C. Given wavelength $\lambda = 6000 \mathring{A}$. Slit separation $d = 5 cm = 5 \times 10^{-2} m$. Distance to screen $D = 50 cm = 0.5 m$. The resultant intensity at point P is equal to the intensity due to individual slits. Let $I_0$ be the intensity due to a single slit. Then the resultant intensity $I = 4 I_0 \cos^2(\phi/2)$, where $\phi$ is the phase difference. Given $I = I_0$. So $I_0 = 4 I_0

Q: What are the options for this this exam Physics question?

A. 4000 · B. 3000 · C. 2000 · D. 1000

Question

Q.45. In a Young double slit experiment, the wavelength of incident of incident light is , the separation between slits  and  is 5 cm the distance between slits plane and screen is 50 cm, as shown in the figure below. If the resultant intensity at P is equal to the intensity due to individual slits, the path different between interfering waves is _______ . [Figure: Young's Double Slit Experiment]

FirstInTest · Accepted Answer

Given wavelength $\lambda = 6000 \mathring{A}$. Slit separation $d = 5 cm = 5 	imes 10^{-2} m$. Distance to screen $D = 50 cm = 0.5 m$. The resultant intensity at point P is equal to the intensity due to individual slits. Let $I_0$ be the intensity due to a single slit. Then the resultant intensity $I = 4 I_0 \cos^2(\phi/2)$, where $\phi$ is the phase difference. Given $I = I_0$. So $I_0 = 4 I_0 \cos^2(\phi/2)$. $1 = 4 \cos^2(\phi/2)$. $\cos^2(\phi/2) = 1/4$. $\cos(\phi/2) = \pm 1/2$. This means $\phi/2 = \pi/3$ or $2\pi/3$ or $4\pi/3$ etc. So $\phi = 2\pi/3$ or $4\pi/3$ etc. The path difference $\Delta x = \frac{\lambda}{2\pi} \phi$. If $\phi = 2\pi/3$, then $\Delta x = \frac{\lambda}{2\pi} \frac{2\pi}{3} = \frac{\lambda}{3}$. $\Delta x = \frac{6000 \mathring{A}}{3} = 2000 \mathring{A}$. This matches option (3).

Answer

4000

Answer

3000

Answer

2000

Answer

1000

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