Respuesta :
Answer:
B. The distance between the slits and the screen is halved.
C. The slit width is doubled.
D. A green, rather than red, light source is used.
E. The experiment is conducted in a water-filled tank.
Explanation:
As we know that the position of first minimum is given as
[tex]a sin\theta = N\lambda[/tex]
so we have
[tex]\theta = sin^{-1}(\fracN\lambda}{a})[/tex]
so width of minimum is given as
[tex]w = L\times sin^{-1}(\fracN\lambda}{a})[/tex]
now if we need to decrease the angular position of minimum
1). so we can decrease the distance of screen from the slit
2). we can decrease the wavelength
3). We can increase the width of the slit
So correct answer will be
B. The distance between the slits and the screen is halved.
C. The slit width is doubled.
D. A green, rather than red, light source is used.
E. The experiment is conducted in a water-filled tank.
These alterations decreases the angles at which the diffraction minima appear is :
B. The distance between the slits and the screen is halved.
C. The slit width is doubled.
D. A green, rather than red, light source is used.
E. The experiment is conducted in a water-filled tank.
"Diffraction angle"
B. The distance between the slits and the screen is halved.
- When slit width is halved angle increases.
- Diffraction angle does not depend on the distance between the slits and the screen.
C. The slit width is doubled.
- The slit width is doubled, angle decreases.
D. A green, rather than red, light source is used.
- As green is having less wavelength than the red therefore when you use green light rather than the red diffraction decreases.
E. The experiment is conducted in a water-filled tank.
- If the experiment is conducted in a water-filled tank, wavelength of the light decreases.
Therefore, the diffraction angle decreases.
Learn more about "diffraction":
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