Calculating Diffraction Angle and Grating Density for Interference Problems

[mustang]

Problem 3. A pair of narrow parallel slits separated by a distance of 0.274 mm are illuminated by the green component from a mercury vapor lamp (wavelength=545.5nm).
What is the angle from the central maximum to the first bright fringe on wither side of the central maximum? Answer in degrees.
Note: If the formula is: d(sin thetha)=m(wavelength)
What do i do next??

Problem 25.
A diffraction grating is calibrated by using the 546.1 m line of mercury vapor. The first-order maximum is found at an angle of 26.09 degrees. Calculate the number of lines per centimeter on this grating. Answer in units of lines/cm.
Note: How do I start?

 

[gnome]

Problem 3:

You have the equation:
d \sin \theta = m \lambda

You have \lambda = 545.5 nm

You have d = 274 mm

The only additional fact you need is that the first bright fringe occurs where m=1.

Solve for \theta.


Problem 25.

The same equation applies. Now you have to solve for d. d = the distance (center-to-center) between slits (lines). Once you know d you can find the number of lines/cm.

 

[mustang]

Regards on problem 3

So the problem is set up like this:
d(sin thetha)=m(wavelength)
274(sin thetha)=1(545.5)
(sin thetha)= 545.5/274
thetha= sin-1(1.9908)
Right?