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IBDP Physics- C.3 Wave phenomena- IB Style Questions For HL Paper 1A -FA 2025

IBDP Physics HL Paper 1 -FA 2025- All Chapters

Question 

In a double-slit experiment using coherent light of wavelength \(\lambda\), the central bright fringe is observed on a screen at point P. A point of destructive interference occurs at point Q. Only one point of constructive interference is observed between P and Q.

What is the path difference at Q?

(A) \(\dfrac{\lambda}{2}\)
(B) \(\lambda\)
(C) \(\dfrac{3\lambda}{2}\)
(D) \(2\lambda\)
▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{C}} \)

For constructive interference, the path difference is

\(\Delta x=n\lambda\)

For destructive interference, the path difference is

\(\Delta x=\left(n+\dfrac{1}{2}\right)\lambda\)

At the central bright fringe P, the path difference is \(0\).

Moving from P towards Q, the first constructive interference occurs at a path difference of \(\lambda\).

Since there is only one constructive interference point between P and Q, Q must correspond to the next destructive interference point:

\(\Delta x=\dfrac{3\lambda}{2}\)

Hence, the correct answer is \( \boxed{\mathrm{C}} \).

Question 

Monochromatic coherent light is directed normally at a diffraction grating with \(N\) slits per unit length. Nine maxima in intensity are produced on a screen far away.

What change will occur to the angle between successive maxima and the number of observable maxima when the number of slits per unit length is changed to \(2N\)?

 Angle between successive maximaNumber of observable maxima
(A)increasedecrease
(B)increaseincrease
(C)decreasedecrease
(D)decreaseincrease
▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{A}} \)

For a diffraction grating, the principal maxima satisfy

\(d\sin\theta=n\lambda\)

where \(d\) is the separation between adjacent slits.

The number of slits per unit length changes from \(N\) to \(2N\). Therefore, the slit spacing becomes smaller:

\(d=\dfrac{1}{N}\)

and

\(d’=\dfrac{1}{2N}=\dfrac{d}{2}\)

For the same order \(n\), a smaller \(d\) requires a larger value of \(\sin\theta\). Hence, the angle between successive maxima increases.

The maximum possible order is limited by

\(n\lambda\leq d\)

Since \(d\) is halved, the maximum observable order is also reduced. Therefore, fewer maxima can be observed.

Thus, the angle between successive maxima increases while the number of observable maxima decreases.

Hence, the correct answer is \( \boxed{\mathrm{A}} \).

Question

Unpolarized light of intensity \( I \) and amplitude \( A \) passes through a polarizing filter.
What is the amplitude of the light transmitted by the polarizing filter?
(A) \( \frac{A^2}{2} \)
(B) \( \frac{A}{4} \)
(C) \( \frac{A}{2} \)
(D) \( \frac{A}{\sqrt{2}} \)
▶️ Answer / Explanation
Detailed solution

When unpolarized light passes through a single polarizing filter, its intensity is reduced to half of its original value:

\( I’ = \frac{I}{2} \)

Intensity is proportional to the square of the amplitude:

\( I \propto A^2 \)

Therefore, for the transmitted wave:

\( \frac{I’}{I} = \frac{(A’)^2}{A^2} = \frac{1}{2} \)

Taking the square root:

\( A’ = \frac{A}{\sqrt{2}} \)

✅ Answer: (D)

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