Question

A two-slit interference experiment is set up and the fringes are displayed on a screen. Then the whole apparatus is immersed in the nearest swimming pool. How does the fringe pattern change?

	a.	The maxima are farther apart due to the resulting decrease in the wavelength of light.
	b.	The maxima are closer together due to the resulting decrease in the wavelength of light.
	c.	The fringe pattern remains the same regardless of the medium.
	d.	The maxima are farther apart due to the resulting increase in the wavelength of light.

A quantity of ideal gas expands at constant pressure of 4.00 x 10^4 Pa. Its volume changes from 2.00 x 10^-3 m^3 to 8.00 x 10^-3 m^3. What is the change in the internal energy of the gas?

	a.	not enough information
	b.	0.0000 J
	c.	-400.0000 J
	d.	-240.0000 J

Answer 1

1) b.The maxima are closer together due to the resulting decrease in the wavelength of light.

because,
distance between two adjacent maxima in air = lamda*R/d
where, lamda --> wavelegnth
d --> slit separation
R --> distance between slits ann screen

distance between two adjacent maxima in water = (lamda*R/d)/n

for water, n = 1.33

so, in water The maxima are closer together due to the resulting decrease in the wavelength of light.

2) The answer is not listed.

change in internal energy = (3/2)*n*R*(T2 - T1)

= (3/2)*P*(V2 - V1) (since P*V = n*R*T)

= (3/2)*4*10^4*(8*10^-3 - 2*10^-3)

= 360 J

Please check your options. The answer must be 360 J

please comment for any further clarification.