What would happen to the distance between the bright fringes produced by a diffraction grating if the entire interference apparatus (light source, grating, and screen) were immersed in water

Answers

Answer 1

Answer:

Since the wavelength would be reduced by the factor of N where N is the index of refraction, fringes would be be closer to one another.

(Similar to the difference of fringes of red light and blue light.)

Related Questions

A nucleus with mass number 229 emits a 3.443 MeV alpha particle. Calculate the disintegration energy Q for this process, taking the recoil energy of the residual nucleus into account.

Answers

Answer:

3.504 MeV

Explanation:

Given that;[tex]\frac{A}{Z} X ---->\frac{A-4}{Z-2} Y +\alpha + Q[/tex]

Also;

[tex]Q= KE_{\alpha } (M_{Y} + M_{\alpha } /M_{Y[/tex])

Mass number of X = 229

Mass number of Y = 225

Mass number of alpha particles = 4

Kinetic energy of alpha particles = 3.443 MeV

Q = 3.443 MeV (225 + 4/225)

Q= 3.504 MeV

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