A plane transmission grating has 15000 lines per inch. Find (a) the resolving power ofthe grating, and (b) the smallest wavelength difference that can be resolved with alight of wavelength 6000 Å in the second order.

(a) The resolving power of a grating is given by the formula:

R = mN

where m is the order of the spectrum and N is the total number of lines in the grating.

Given that the grating has 15000 lines per inch, we need to convert this to lines per meter.

1 inch = 2.54 cm = 0.0254 m

So, N = 15000 lines/inch * 1/0.0254 m/inch = 590551.18 lines/m

For the maximum resolving power, we consider the highest order of spectrum (m) that can be seen. This is typically around the value of m = 2.

So, R = mN = 2 * 590551.18 = 1181102.36

(b) The smallest wavelength difference that can be resolved (Δλ) is given by the formula:

Δλ = λ/R

where λ is the wavelength of light.

Given that the wavelength of light is 6000 Å, we need to convert this to meters.

1 Å = 10^-10 m

So, λ = 6000 Å * 10^-10 m/Å = 6 * 10^-7 m

So, Δλ = λ/R = 6 * 10^-7 m / 1181102.36 = 5.08 * 10^-13 m

Converting back to Å,

Δλ = 5.08 * 10^-13 m * 10^10 Å/m = 0.00508 Å

So, the smallest wavelength difference that can be resolved is 0.00508 Å.