AP Physics Practice Questions: Optics

AP Physics multiple choice questions on geometric and physical optics. Each question includes detailed solutions at the bottom of the page.

1. Light travels from air (n=1.00) into glass (n=1.50) at an angle of incidence of 30°. What is the angle of refraction?

   A) 15.7°
   B) 19.5°
   C) 30.0°
   D) 48.6°
   E) 60.0°

2. A concave mirror has focal length 20 cm. Where should an object be placed to form a real image twice the object's size?

   A) 10 cm
   B) 15 cm
   C) 20 cm
   D) 30 cm
   E) 40 cm

3. A converging lens (f=15 cm) forms an image of an object placed 30 cm from the lens. What is the image distance?

   A) 10 cm
   B) 15 cm
   C) 30 cm
   D) 40 cm
   E) 60 cm

4. Light of wavelength 600 nm passes through a single slit of width 0.1 mm onto a screen 1 m away. What is the width of the central maximum?

   A) 12 mm
   B) 24 mm
   C) 36 mm
   D) 48 mm
   E) 60 mm

5. What is the critical angle for light going from diamond (n=2.42) to air (n=1.00)?

   A) 22.1°
   B) 24.4°
   C) 30.0°
   D) 45.0°
   E) 60.0°

Solutions

1. B) 19.5°
   \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \)
   \( 1.00 \times \sin 30^\circ = 1.50 \times \sin \theta_2 \)
   \( 0.5 = 1.50 \sin \theta_2 \)
   \( \sin \theta_2 = \dfrac{0.5}{1.5} = \dfrac{1}{3} \approx 0.3333 \)
   \( \theta_2 = \sin^{-1}(0.3333) \approx 19.47^\circ \)

2. D) 30 cm
   For concave mirror, real image with magnification \(-2\) (negative for real image)
   \( m = -\dfrac{i}{o} = -2 \Rightarrow i = 2o \)
   \( \dfrac{1}{f} = \dfrac{1}{o} + \dfrac{1}{i} = \dfrac{1}{o} + \dfrac{1}{2o} = \dfrac{3}{2o} \)
   \( \dfrac{1}{20} = \dfrac{3}{2o} \Rightarrow o = 30 \,\text{cm} \)

3. C) 30 cm
   Thin lens equation: \( \dfrac{1}{f} = \dfrac{1}{o} + \dfrac{1}{i} \)
   \( \dfrac{1}{15} = \dfrac{1}{30} + \dfrac{1}{i} \)
   \( \dfrac{1}{i} = \dfrac{1}{15} - \dfrac{1}{30} = \dfrac{2}{30} - \dfrac{1}{30} = \dfrac{1}{30} \)
   \( i = 30 \,\text{cm} \)

4. A) 12 mm
   Single slit diffraction: width of central maximum \( = \dfrac{2\lambda L}{a} \)
   \( = \dfrac{2 \times 600 \times 10^{-9} \times 1}{0.1 \times 10^{-3}} \)
   \( = \dfrac{1200 \times 10^{-9}}{10^{-4}} = 1200 \times 10^{-5} = 0.012 \,\text{m} = 12 \,\text{mm} \)

5. B) 24.4°
   \( \sin \theta_c = \dfrac{n_2}{n_1} = \dfrac{1.00}{2.42} \approx 0.4132 \)
   \( \theta_c = \sin^{-1}(0.4132) \approx 24.4^\circ \)