Unit 13: Geometric and Physical Optics
Weightage : 15–18%
An experiment is conducted to determine the critical angle for light going from glass into air, as shown above. A linear plot is made with a vertical axis of \( \sin(\theta_{1}) \) and a horizontal axis of \( \sin(\theta_{2}) \). How is the critical angle determined from the graph?
Correct Answer: \( \boxed{\mathrm{C}} \)
Snell’s law for light traveling from glass into air is
\( n_{\mathrm{glass}}\sin(\theta_{1})=n_{\mathrm{air}}\sin(\theta_{2}) \)
The critical angle occurs when the refracted ray travels along the boundary, so
\( \theta_{2}=90^\circ \)
Since
\( \sin(90^\circ)=1 \),
the critical angle corresponds to the point on the graph where the vertical-axis value, \( \sin(\theta_{1}) \), reaches its maximum value of \(1\). The corresponding value on the horizontal axis, \( \sin(\theta_{2}) \), gives \( \sin(\theta_{\mathrm{c}}) \), from which the critical angle is determined.
Therefore, the critical angle is obtained from the horizontal-axis value corresponding to the maximum vertical value.
Thus, the correct answer is (C).
When a light wave passes from air into water, what properties of the wave change?
Correct Answer: \( \boxed{\mathrm{D}} \)
When light enters a different medium, its frequency remains constant, but its speed changes according to
\( v=\frac{c}{n}. \)
Since water has a higher refractive index than air, the speed of light decreases as it enters the water.
Although the wavelength also changes according to
\( v=f\lambda, \)
the question asks for a single correct answer. The fundamental property that changes due to the change in medium is the speed.
Therefore, the correct answer is (D).
A student performs an experiment to determine the index of refraction of a plastic block. The student shines a laser beam from air \((n_{\mathrm{air}}=1)\) into a piece of plastic at an angle of incidence \(\theta_1\) and measures the angle of refraction \(\theta_2\), as shown in Figure 1 above. Measurements are taken as \(\theta_1\) is increased. From the data, the student produces the two graphs shown above.
Which of the following is equal to the index of refraction of the plastic block?
Correct Answer: \( \boxed{\mathrm{D}} \)
The relationship between the incident and refracted angles is given by Snell’s law:
\( n_{\mathrm{air}}\sin\theta_1=n_{\mathrm{plastic}}\sin\theta_2. \)
Since \(n_{\mathrm{air}}=1\),
\( \sin\theta_1=n_{\mathrm{plastic}}\sin\theta_2. \)
Rearranging,
\( \sin\theta_2=\dfrac{1}{n_{\mathrm{plastic}}}\sin\theta_1. \)
Graph 2 plots \( \sin\theta_2 \) versus \( \sin\theta_1 \), giving a straight-line equation of the form
\( y=mx, \)
where
\( m=\dfrac{1}{n_{\mathrm{plastic}}}. \)
Therefore,
\( n_{\mathrm{plastic}}=\dfrac{1}{\text{slope}}. \)
Graph 1 is not linear, so its slope does not directly represent the refractive index.
Hence, the index of refraction of the plastic is equal to the inverse of the slope of the best-fit line for Graph 2.
