And the angle from the mirror up to the ray is ϕ so

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and the angle from the mirror up to the ray is ϕ so we can say 30° = θ + ϕ . On the right, the angle from the reflected ray to the mirror is ϕ ; the angle from the horizontal up to the ray is 10° (it’s a corresponding angle to the 10° on the right) and the angle from the horizontal down to the mirror is θ (it’s an alternate interior angle of the original θ ) so we can say ϕ = 10° + θ . Combining these equations lets us solve for θ . 17. If the depth of the water is w , then the light travels a vertical distance of h – w in the air. We can relate this to the horizontal distance it travels in the air, x a , through: a x w h = ) tan( θ or x a = ( h – w ) tan( θ ). We can use Snell’s Law to find θ w , the angle of refraction (22.1°). We can relate the depth of the water to this angle and the horizontal distance the light travels in the water, x w , through: w x w w = ) tan( θ or x w = w tan( θ w ). The two horizontal distances add to 25 cm (25 cm = x a + x w ) so we can solve for the water’s depth. 18. We can eliminate B, D, and E, since converging lenses do create enlarged upright images when the object is closer to the lens than the focal point (so f > 4 cm). The magnification is 1.667 so we can find the image distance (–6.668 cm). Using this, the object distance, and the thin lens equations gives us A. 19. We can sketch the ray diagrams or use the thin lens equation to get the order. 20. We can eliminate all but D and E (A and B since the frequency of light doesn’t change when it travels through different media, C since the beams will have different wavelengths in vacuum). In a vacuum, 2 has a wavelength longer than 530 nm (light’s wavelength always decreases when it travels through a non-vacuum medium); this means in vacuum 1 has a smaller wavelength so it must have a higher frequency.
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