This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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b) Step 1: Identify the given values. The real depth of the fish, d_real = 3.6 \, m. The refractive index of water, n = (4)/(3). Step 2: Use the formula for apparent depth. The relationship between real depth, apparent depth, and refractive index is given by: d_apparent = d_realn Step 3: Substitute the values and calculate the apparent depth. d_apparent = 3.6 \, m(4)/(3) d_apparent = 3.6 \, m × (3)/(4) d_apparent = (10.8)/(4) \, m d_apparent = 2.7 \, m The apparent depth of the fish is 2.7 m. c) i) Step 1: Identify the angles from the first figure. Angle of incidence in air, i = 42^. Angle of refraction in water, r = 30^. The refractive index of air, n_air = 1. Step 2: Apply Snell's Law. Snell's Law states n_air i = n_water r. 1 × 42^ = n_water × 30^ Step 3: Solve for the refractive index of water, n_water. n_water = ( 42^)/( 30^) n_water = (0.6691)/(0.5) n_water = 1.3382 The refractive index of water is 1.34 (rounded to two decimal places). c) ii) Step 1: Identify the angles from the second figure and use the calculated refractive index. Angle of refraction in water, r = 20^. Angle of incidence in air, i = . Refractive index of air, n_air = 1. Refractive index of water, n_water = 1.3382 (from part c i). Step 2: Apply Snell's Law. n_air = n_water r 1 × = 1.3382 × 20^ Step 3: Solve for angle . = 1.3382 × 0.3420 = 0.4576 = (0.4576) = 27.23^ Angle is 27.2^ (rounded to one decimal place). 18. (a) Two conditions for total internal reflection to occur are: • Light must travel from an optically denser medium to an optically rarer medium. • The angle of incidence in the denser medium must be greater than the critical angle for the interface between the two media. That's 2 down. 3 left today — send the next one.