Parallel rays of green mercury light with a wavelength of 546 nm pass through a slit covering a lens with a focal length of 60.0 cm. In the focal plane of the lens the distance from the central maximum to the first minimum is 10.2 mm. What is the width of the slit
a/2 sin theta = lambda / 2
The Attempt at a Solution
Tangent theta is opposite over adjacent... The adjacent, I think, is the focal length, 60.0 cm. The opposite is clearly 10.2 mm. So, tan theta = 10.2 mm / 600 mm. Then to get a, the width of the slit, I use the above equation.
I am not familiar with lenses. So, I am not sure if the "adjacent" is correct.