Answer
400 nm and 600 nm.
Work Step by Step
The brightness maximum for the wavelength of 1200 nm occurs at a certain angle.
Solve equation 24–4 for the slit separation at the first-order 1200 nm line, and at a different wavelength.
$$d=\frac{m_1\lambda_1}{sin\theta_1}=\frac{1(1200nm)}{ sin\theta}$$
$$d=\frac{m_2\lambda_2}{sin\theta}$$
The same diffraction grating, with the same slit spacing d, is used for both experiments. We equate the two expressions. At this same angle, a bright band may occur for another wavelength.
$$\frac{1(1200nm)}{ sin\theta}=\frac{m_2\lambda_2}{sin\theta}$$
Solve for the unknown wavelength if m = 2.
$$1200nm=2\lambda$$
$$\lambda=600nm$$
Solve for the unknown wavelength if m = 3.
$$1200nm=3\lambda$$
$$\lambda=400nm$$
Solve for the unknown wavelength if m = 4.
$$1200nm=4\lambda$$
$$\lambda=300nm$$
If m = 5, 6, etc. the wavelengths will be even shorter. In the specified range from 360 nm to 2000 nm, the 2 wavelengths that have a brightness maximum at the angle where 1200 nm radiation has a brightness maximum are 400 nm and 600 nm.
