# 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           $\frac{sin30^\cdot&space;}{sin45^\cdot&space;}$

B           $\frac{sin45^\cdot&space;}{sin30^\cdot&space;}$

C           $\frac{sin45^\cdot&space;}{sin60^\cdot&space;}$

D          $\frac{sin60^\cdot&space;}{sin45^\cdot&space;}$

Answer/Explanation

Ans: B

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

### 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

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