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Step-by-Step Solution
Step 1: Identify the Key Variables
Let the number of fringes observed when the wavelength of light is $ \lambda_1 $ be $ N_1 $. Similarly, let the number of fringes for another wavelength $ \lambda_2 $ be $ N_2 $. Also, let the distance between the slits and the screen be D, and the separation between the slits be d.
Step 2: Express the Fringe Width
The fringe width ($ w $) for Young’s double slit experiment is given by:
$$ w = \frac{\lambda D}{d}. $$
If $ N $ fringes occupy a segment of length $ l $ on the screen, then
$$ N \times w = l. $$
Step 3: Apply the Condition for the Same Segment Length
The same segment length $ l $ is used for two different wavelengths $ \lambda_1 $ and $ \lambda_2 $. Hence, for the first case,
$$ N_1 \times \frac{\lambda_1 D}{d} = l, $$
and for the second case,
$$ N_2 \times \frac{\lambda_2 D}{d} = l. $$
Since the segment length $ l $ is the same in both cases, equate the two expressions:
$$ N_1 \times \frac{\lambda_1 D}{d} = N_2 \times \frac{\lambda_2 D}{d}. $$
Step 4: Simplify the Equation
Canceling the common factor $ \frac{D}{d} $, we get:
$$ N_1 \, \lambda_1 = N_2 \, \lambda_2. $$
Step 5: Substitute the Known Values
Given:
$$ N_1 = 16, \quad \lambda_1 = 700 \,\text{nm}, \quad \lambda_2 = 400 \,\text{nm}. $$
So,
$$ 16 \times 700 = N_2 \times 400. $$
Step 6: Solve for $ N_2 $
$$ 16 \times 700 = N_2 \times 400, $$
$$ 11200 = 400 \, N_2, $$
$$ N_2 = \frac{11200}{400} = 28. $$
Step 7: State the Final Answer
The number of fringes observed in the same segment of the screen when the wavelength is changed to 400 nm is 28.
