1. **Problem:** Calculate the speed of a wave with frequency 500 Hz and wavelength 0.6 m. Formula: Wave speed $v$ is given by $$v = f \times \lambda$$ where $f$ is frequency and $\lambda$ is wavelength. Calculation: $$v = 500 \times 0.6 = 300\ \text{m/s}$$ 2. **Problem:** Find the wavelength of a wave with speed 340 m/s and frequency 170 Hz. Formula: $$\lambda = \frac{v}{f}$$ Calculation: $$\lambda = \frac{340}{170} = 2\ \text{m}$$ 3. **Problem:** Define amplitude. Answer: Amplitude is the maximum displacement of particles in the medium from their rest position during wave motion. It relates to the energy and loudness of the wave. 4. **Problem:** Name the phenomenon where a coin appears closer to the surface in water. Answer: Refraction. 5. **Problem:** Explain why the coin appears raised. Explanation: Light bends when it passes from water to air due to change in speed, making the coin appear at a shallower depth than it actually is. 6. **Problem:** State two practical applications of refraction. Answer: Lenses in eyeglasses and cameras, and the apparent bending of a straw in a glass of water. 7. **Problem:** Find the refractive index of glass where speed of light is 183,000 km/s. Formula: $$n = \frac{c}{v}$$ where $c = 3 \times 10^8$ m/s (speed of light in vacuum), $v$ is speed in medium. Calculation: Convert $v$ to m/s: $$183,000\ \text{km/s} = 1.83 \times 10^8\ \text{m/s}$$ $$n = \frac{3 \times 10^8}{1.83 \times 10^8} \approx 1.64$$ 8. **Problem:** Find speed of light in gasoline with refractive index 1.40. Formula: $$v = \frac{c}{n}$$ Calculation: $$v = \frac{3 \times 10^8}{1.40} \approx 2.14 \times 10^8\ \text{m/s}$$ 9. **Problem:** For a concave mirror with focal length $f=15.2$ cm, object height $h_o=4.00$ cm, object distance $d_o=45.7$ cm, find image distance $d_i$ and image height $h_i$. Formula (mirror equation): $$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$$ Calculate $d_i$: $$\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} = \frac{1}{15.2} - \frac{1}{45.7}$$ Calculate each term: $$\frac{1}{15.2} \approx 0.06579, \quad \frac{1}{45.7} \approx 0.02188$$ $$\frac{1}{d_i} = 0.06579 - 0.02188 = 0.04391$$ $$d_i = \frac{1}{0.04391} \approx 22.78\ \text{cm}$$ Magnification $m$: $$m = -\frac{d_i}{d_o} = -\frac{22.78}{45.7} \approx -0.498$$ Image height: $$h_i = m \times h_o = -0.498 \times 4.00 = -1.99\ \text{cm}$$ Negative sign means image is inverted. 10. **Problem:** State laws of reflection. Answer: - The angle of incidence equals the angle of reflection ($i = r$). - The incident ray, reflected ray, and normal lie in the same plane. 11. **Problem:** Calculate angle of refraction when light passes from air ($n=1$) into glass ($n=1.5$) at incidence angle $i=30^\circ$. Snell's Law: $$n_1 \sin i = n_2 \sin r$$ Calculate $r$: $$\sin r = \frac{n_1}{n_2} \sin i = \frac{1}{1.5} \times \sin 30^\circ = \frac{1}{1.5} \times 0.5 = 0.3333$$ $$r = \sin^{-1}(0.3333) \approx 19.47^\circ$$ 12. **Problem:** Calculate critical angle for glass ($n=1.5$) to air ($n=1$). Formula: $$\sin \theta_c = \frac{n_2}{n_1}$$ Calculation: $$\sin \theta_c = \frac{1}{1.5} = 0.6667$$ $$\theta_c = \sin^{-1}(0.6667) \approx 41.81^\circ$$ 13. **Problem:** State one practical application of total internal reflection. Answer: Optical fibers for communication. 14. **Problem:** Explain why total internal reflection cannot occur from air to glass. Explanation: Total internal reflection requires light to travel from a denser to a rarer medium. Air is less dense than glass, so light going from air to glass cannot undergo total internal reflection. 15. **Multiple Choice Answers:** 1. B 2. B 3. C 4. C 5. B 6. B 7. C 8. B 9. B 10. A 14. C 15. C 16. C 17. A 18. B 19. A 20. A 21. D 22. B 23. B 24. C 25. A **Note:** For question 11, angle of incidence and reflection are equal; if angle A is incidence, angle B is reflection (diagram not provided). **Summary:** All calculations and explanations are shown step-by-step for clarity and learning.