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Question #43735

Difficulty: 8/10 | 🟢 Free

The Question

In a single-slit diffraction experiment, the wavelength of the incident light is given as $\lambda = (600.0 \pm 6.0) \text{ nm}$ . The width of the slit, $a$ , is measured using a screw gauge with a pitch of $0.5 \text{ mm}$ and $50$ divisions on its circular scale. During the measurement of the slit width, the main scale reading is $0 \text{ mm}$ and the circular scale reading is $25$ divisions. Assuming the screw gauge has no zero error, what is the maximum percentage error in the calculated angular position of the second diffraction minimum? (Assume the small-angle approximation is valid).

Options

A.
$5\%$
B.
$6\%$
C.
$3\%$
D.
$4\%$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Single Slit Diffraction Minima|Relative Error Propagation in Powers|Calculus/Approximation

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to propagate experimental uncertainties through a quotient relationship derived from the single-slit diffraction formula.

A common trap is incorrectly applying the power rule to the constant $n=2$ (leading to a $6\%$ error by erroneously doubling the wavelength's uncertainty) or incorrectly subtracting the relative errors of the numerator and denominator (yielding $3\%$ ).

The correct physical framework requires recognizing that exact constants do not contribute to relative error and that maximum relative uncertainties must always be added to account for the worst-case error bounds.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Wave Optics - Class 12
Concept	Single Slit Diffraction Minima
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free