Question

A double slit is illuminated by light of wavelengths 6000 Å. The slits are 0.1 cm apart and screen is placed 1.5 m. away. Calculate the angular position of 9th maximum in radian.

Answer 1

λ = 6000 Å = 6 x 10^-7 m, d = 0.1 cm = 10^-3 m, D = 1.5 m, n = 9

Angular position of nth maximum

θ_n = \(\frac{yn}{D} = \frac{\frac{n\lambda D}{d}}{D} = \frac{n\lambda}{d}\)

or θ_n = \(\frac{9 \times 6 \times 10^{-7}}{10^{-3}}\)

= 54 x 10^-4

= 0.0054 radian.

Answer 2

To solve the problem, we need to calculate the angular position of the 9th maximum in a double-slit interference pattern. We'll use the formula for the angular position of the maxima:

\[ \theta_n = \frac{n \lambda}{d} \]

where:

- \( \theta_n \) is the angular position of the nth maximum.

- \( n \) is the order of the maximum (for the 9th maximum, \( n = 9 \)).

- \( \lambda \) is the wavelength of the light.

- \( d \) is the distance between the slits.

Given:

- \( \lambda = 6000 \) Å = \( 6000 \times 10^{-10} \) m

- \( d = 0.1 \) cm = \( 0.1 \times 10^{-2} \) m = \( 1 \times 10^{-3} \) m

- \( n = 9 \)

Now, we can substitute these values into the formula:

\[ \theta_9 = \frac{9 \times 6000 \times 10^{-10} \text{ m}}{1 \times 10^{-3} \text{ m}} \]

\[ \theta_9 = \frac{9 \times 6000 \times 10^{-10}}{1 \times 10^{-3}} \]

\[ \theta_9 = \frac{54000 \times 10^{-10}}{10^{-3}} \]

\[ \theta_9 = 54000 \times 10^{-7} \]

\[ \theta_9 = 5.4 \times 10^{-3} \text{ radians} \]

So, the angular position of the 9th maximum is \( 5.4 \times 10^{-3} \) radians.