Question

36.5 diffraction gratings
a diffraction grating that is 3.11 cm wide produces a second order intensity maximum at 19.0° when light of wavelength 675 nm is used. what must be the total number of lines on the grating?
lines
save for later\t\tsubmit answer

Explanation:

Step1: Recall Diffraction Grating Formula

The formula for diffraction grating is \( d\sin\theta = m\lambda \), where \( d \) is the distance between adjacent lines, \( \theta \) is the angle of diffraction, \( m \) is the order of the maximum, and \( \lambda \) is the wavelength of light. We need to find the total number of lines \( N \), and we know that \( d=\frac{w}{N} \), where \( w \) is the width of the grating.

Step2: Rearrange Formula to Solve for \( N \)

Substitute \( d = \frac{w}{N} \) into the diffraction formula: \( \frac{w}{N}\sin\theta = m\lambda \). Rearranging for \( N \), we get \( N=\frac{w\sin\theta}{m\lambda} \).

Step3: Convert Units

First, convert the width \( w = 3.11\space cm=3.11\times 10^{-2}\space m \), wavelength \( \lambda = 675\space nm = 675\times 10^{-9}\space m \), \( m = 2 \), \( \theta=19.0^\circ \).

Step4: Plug in Values

Calculate \( \sin(19.0^\circ)\approx\sin(19^\circ)\approx 0.3256 \). Then substitute into the formula:
\( N=\frac{3.11\times 10^{-2}\space m\times0.3256}{2\times675\times 10^{-9}\space m} \)

Step5: Perform Calculation

First, calculate the numerator: \( 3.11\times 10^{-2}\times0.3256\approx 3.11\times0.3256\times 10^{-2}\approx1.0126\times 10^{-2} \)
Denominator: \( 2\times675\times 10^{-9}=1350\times 10^{-9}=1.35\times 10^{-6} \)
Then \( N=\frac{1.0126\times 10^{-2}}{1.35\times 10^{-6}}=\frac{1.0126}{1.35}\times 10^{4}\approx0.7501\times 10^{4}=7501 \) (approximate value, more precise calculation: \( \frac{3.11\times 10^{-2}\times\sin(19^\circ)}{2\times675\times 10^{-9}} \))

Let's do a more precise calculation:
\( \sin(19.0^\circ)=\sin(19^\circ)\approx0.3255681545 \)
Numerator: \( 3.11\times 10^{-2}\times0.3255681545 = 3.11\times0.3255681545\times 10^{-2}\approx1.012517\times 10^{-2} \)
Denominator: \( 2\times675\times 10^{-9}=1350\times 10^{-9}=1.35\times 10^{-6} \)
\( N=\frac{1.012517\times 10^{-2}}{1.35\times 10^{-6}}=\frac{1.012517}{1.35}\times 10^{4}\approx0.7500126\times 10^{4}=7500.126\approx7500 \) (or more accurately, let's compute step by step)
\( N=\frac{3.11\times 10^{-2}\times\sin(19.0^\circ)}{2\times675\times 10^{-9}} \)
\(=\frac{3.11\times 10^{-2}\times0.325568}{2\times675\times 10^{-9}} \)
\(=\frac{3.11\times0.325568\times 10^{-2}}{1350\times 10^{-9}} \)
\(=\frac{3.11\times0.325568}{1350}\times 10^{7} \)
\( 3.11\times0.325568\approx1.0125 \)
\( \frac{1.0125}{1350}\approx0.00075 \)
\( 0.00075\times 10^{7}=7500 \)

Answer:

\( \boxed{7500} \) (approximate, more precise calculation may give around 7500 - 7510, but 7500 is a reasonable answer)