A diffraction grating has 600 lines per mm and a screen is 2.4 m away. How wide is the second-order visible spectrum on the screen (from 400 nm to 700 nm)?

What “width of the second-order spectrum” means

A diffraction grating spreads different wavelengths to different angles. The spectrum width (for one side of the central maximum) is the distance on the screen between where the shortest visible wavelength lands and where the longest visible wavelength lands, for the same diffraction order.

Convert the grating ruling to slit spacing $d$

600 lines/mm means: $$600\,\text{lines/mm} = 600\times 10^3\,\text{lines/m}$$ So the spacing is $$d=\frac{1}{600\times10^3}=1.6667\times10^{-6}\,\text{m}$$

Use the grating equation for second order

The grating condition is $$d\,\sin\theta = m\lambda$$ For second order, $m=2$.

For violet end ($\lambda_1=400\,\text{nm}$): $$\sin\theta_1 = \frac{2(400\times10^{-9})}{1.6667\times10^{-6}}=0.48 \Rightarrow \theta_1\approx 28.7^\circ$$

For red end ($\lambda_2=700\,\text{nm}$): $$\sin\theta_2 = \frac{2(700\times10^{-9})}{1.6667\times10^{-6}}=0.84 \Rightarrow \theta_2\approx 57.0^\circ$$

Convert angles to screen positions and subtract

With screen distance $L=2.4\,\text{m}$, the displacement from the center is $$y=L\tan\theta$$ So $$y_1=2.4\tan(28.7^\circ)\approx 2.4(0.546)\approx 1.31\,\text{m}$$ $$y_2=2.4\tan(57.0^\circ)\approx 2.4(1.539)\approx 3.69\,\text{m}$$ Width of the second-order visible spectrum (one side) is $$\Delta y=y_2-y_1\approx 3.69-1.31=2.38\,\text{m}$$

Quick check

The larger wavelength should appear farther from the center because it diffracts to a larger angle, and $3.69\,\text{m} > 1.31\,\text{m}$, which matches that expectation.