Question

light of wavelength 542 nm and 813 nm are incident on a diffraction grating with 4120 lines/cm. for what order would the 813 nm light have the same angle as the 542 nm light of the next higher order?
hint
the diffraction pattern is given by
$d\sin(\theta) = m\lambda$,
and we are looking for $m$ such that
$\theta_1 = \theta_2$
$=> m_1\lambda_1 = m_2\lambda_2$,
where $m_1 = m$ and $m_2 = (m+1).$

Explanation:

Step1: Identify given values

Let \( \lambda_1 = 813 \, \text{nm} \), \( \lambda_2 = 542 \, \text{nm} \), \( m_1 = m \), \( m_2 = m + 1 \). From the diffraction formula \( d\sin\theta = m\lambda \), and since \( \theta_1=\theta_2 \), we have \( m_1\lambda_1=m_2\lambda_2 \).

Step2: Substitute \( m_1 \) and \( m_2 \)

Substitute \( m_1 = m \) and \( m_2 = m + 1 \) into \( m_1\lambda_1=m_2\lambda_2 \):
\( m \times 813 = (m + 1) \times 542 \)

Step3: Expand and solve for \( m \)

Expand the right - hand side: \( 813m=542m + 542 \)
Subtract \( 542m \) from both sides: \( 813m-542m=542 \)
\( 271m = 542 \)
Divide both sides by 271: \( m=\frac{542}{271}=2 \)

Answer:

\( 2 \)