IB DP Physics Unit 4. Waves- Topic 4.3 – Wave characteristics: IB Style Question Bank HL Paper 1

Question

Wavefronts travel from air to medium Q as shown.

                 

What is the refractive index of Q?

A           \frac{sin30^\cdot }{sin45^\cdot }

B           \frac{sin45^\cdot }{sin30^\cdot }

C           \frac{sin45^\cdot }{sin60^\cdot }

D          \frac{sin60^\cdot }{sin45^\cdot }

Answer/Explanation

Ans: B

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

\(\frac{n_1}{n_2}=\frac{sin \theta_2}{sin \theta_1} \; snell \;law\)
\(\frac{n_{air}}{n_Q}=\frac{sin\; r}{sin\;i}\)
\(\therefore n_Q =n_{air}\times\frac{sin\;i}{sin\; r} =\frac{sin \;45^0}{sin \;30^0}\)
\(\because n_{air} =1\)
\(\therefore n_Q =\frac{sin \;45^0}{sin \;30^0}\)\)

Question

Unpolarized light of intensity I1 is incident on a polarizer. The light that passes through this polarizer then passes through a second polarizer.

       

The second polarizer can be rotated to vary the intensity of the emergent light. What is the maximum value of the intensity emerging from the second polarizer?

A \frac{l_1}{4}

B \frac{l_1}{2}

C \frac{2l_1}{3}

D I_1

Answer/Explanation

Ans: B

\(I_3=I_2cos^2\theta\)
\(But\; I_2 = \frac{I_1}{2}\)
\(I_3=\frac{I_1}{2}cos^2\theta\)
\((I_3)_{max} =\frac{I_1}{2} \because (cos^2\theta)_{max} =1\)

Question

Which graph shows the variation of amplitude with intensity for a wave?

               

Answer/Explanation

Ans: A  (Markscheme is B which is wrong)

We know that Intensity and Amplitude is related as  \(I\propto A^2\)
or
\(I = kA^2\)
This is Similar to
\(y=kx^2 \) (parabola)with symmetry around x axis (Intensity)

\(I=kA^2\)
\(A^2=\frac{I}{k}\)
or
\(A=\pm \sqrt{\frac{I}{k}}\)
So for one I there are 2 A
\(A_1=\sqrt{\frac{I}{k}}\)
\(A_2=-\sqrt{\frac{I}{k}}\)
This is possible in Graph A only
Hence A is correct Answer