IBDP Physics Physics Unit 4.4 – Wave behavior: IB Style Question Bank – HL Paper 1

Question

Monochromatic light of wavelength λ is incident on a double slit. The resulting interference pattern is observed on a screen a distance y from the slits. The distance between consecutive fringes in the pattern is 55 mm when the slit separation is aλ, y and a are all doubled. What is the new distance between consecutive fringes?

A 55 mm

B 110 mm

C 220 mm

D 440 mm

Answer/Explanation

Ans: B

Fringe width is given by
\(\beta =\frac{\lambda \times y}{a}\)
if \(\lambda\) , y and a are doubled
\({\beta }’ =\frac{2\lambda \times2y}{2a}=2 \times \frac{\lambda \times y}{a}=2\times 55=110 \; mm\)

Question

Light is incident from air on the surface of a transparent medium.

When V is equal to the Brewster angle, which angle is equal to 90°?

A.     \({\text{V}} + {\text{W}}\)

B.     W only

C.     \({\text{X}} + {\text{Y}}\)

D.     Z only

Answer/Explanation

Markscheme

C

Ref: https://www.iitianacademy.com/ib-dp-physics-topic-4-waves-4-4-wave-behavior-study-notes/

POLARIZATION BY REFLECTION (BREWSTER’S LAW)

During reflection of a wave, we obtain a particular angle called angle of polarisation, for which the reflected light is completely plane polarised.
\(V+Z=90\)
Now V = W Since Angle of incident = Angle of reflection ( Law of reflection)
Hence
\(W+Z= 90\)
Now \(W+X+Y+Z =180^0\) ( Angle of Straight line)
Hence \( X+Y= 180 -90 =90^0\)