Question

When a system of multiple long narrow slits (width 2µm and period 4µm) is illuminated with a laser of wavelength 600nm. There are 40 minima between the two consecutive principal maxima observed in its diffraction pattern. Then maximum resolving power of the system is _______

Answer 1

Correct Answer: 246

Resolving Power of a Diffraction Grating 

Step 1: Formula for Resolving Power

The resolving power \( R \) of a diffraction grating is given by:

\[ R = nN \]

where:

• \( n \) is the order of diffraction.
• \( N \) is the total number of slits illuminated.

Step 2: Finding the Number of Slits

The number of minima \( m \) between two principal maxima is related to the number of slits \( N \) by:

\[ m = N - 1 \]

Substituting \( m = 40 \):

\[ N = m + 1 = 40 + 1 = 41 \]

Step 3: Slit Separation Calculation

Given:

• Period of the grating: \( d = 4\mu m \).
• Slit width: \( a = 2\mu m \).

The slit separation is:

\[ d - a = 4\mu m - 2\mu m = 2\mu m \]

Step 4: Resolving Power for First-Order Diffraction

The first-order diffraction (\( n = 1 \)) is typically used for maximum resolving power.

Substituting \( n = 1 \) and \( N = 41 \) into the resolving power formula:

\[ R = nN = 1 \times 41 = 41 \]

Step 5: Maximum Order of Diffraction

The maximum order \( n \) for diffraction can be determined using the grating equation:

\[ n\lambda \leq d \]

where:

• \( \lambda = 600nm = 0.6\mu m \).
• \( d = 4\mu m \).

Solving for \( n \):

\[ n \leq \frac{d}{\lambda} = \frac{4}{0.6} \approx 6.67 \]

The maximum integer \( n \) is \( 6 \).

Step 6: Resolving Power for Maximum Order

Substituting \( n = 6 \) and \( N = 41 \) into the resolving power formula:

\[ R = nN = 6 \times 41 = 246 \]

Final Answer

The maximum resolving power of the system is 246.