Calculate the critical angle for total internal reflection at the X-Y boundary given refractive indices n1 = 1.80 and n2 = 1.20.

Here are the solutions to your physics questions: 20. Light travels from medium X (n_1 = 1.80) to medium Y (n_2 = 1.20). Calculate the critical angle for total internal reflection at the X-Y boundary. (3 marks) Step 1: Identify the given refractive indices. Refractive index of medium X, n_1 = 1.80 Refractive index of medium Y, n_2 = 1.20 For total internal reflection, light must travel from a denser medium to a rarer medium, which is the case here (n_1 > n_2). Step 2: Apply the formula for the critical angle. The critical angle (_c) is given by Snell's Law when the angle of refraction is 90^: n_1 _c = n_2 90^ Since 90^ = 1: n_1 _c = n_2 _c = (n_2)/(n_1) Step 3: Substitute the values and calculate _c. _c = (1.20)/(1.80) _c = (2)/(3) _c ≈ 0.6667 _c = (0.6667) _c ≈ 41.81^ The critical angle is 41.81^. 21. With the aid of a clearly labeled diagram, explain how a 45^-45^-90^ glass prism can be used to redirect a ray of light through 90^. Show clearly the angles of incidence and reflection. (4 marks) A 45^-45^-90^ glass prism can redirect a ray of light through 90^ using total internal reflection. Explanation: 1. A ray of light enters one of the shorter faces of the prism perpendicularly. Since it enters normally, it passes into the glass without any deviation (angle of incidence is 0^, so angle of refraction is 0^). 2. This ray then strikes the hypotenuse face of the prism. Due to the geometry of the prism, the angle of incidence at this face is 45^ (measured from the normal to the surface). 3. For typical glass (refractive index n ≈ 1.5), the critical angle for a glass-air interface is approximately 41.8^. Since the angle of incidence (45^) is greater than the critical angle, total internal reflection occurs. 4. The light ray is reflected internally at an angle of reflection of 45^. 5. The reflected ray then strikes the other shorter face of the prism perpendicularly and exits without deviation. 6. The overall effect is that the light ray is redirected by 90^ from its original path. Diagram: ` /|\ / | \ / | \ / | \ / | \ / | \ +------+------+ | | | | | | | | | | | | | | | | | | +------+------+ \ | / \ | / \ | / \ | / \ | / \|/ V ` (Imagine a right-angled isosceles triangle. Light enters horizontally from the left, hits the hypotenuse, reflects downwards, and exits vertically. The angles of incidence and reflection at the hypotenuse are both 45^.) 22. With the aid of a labeled diagram, explain how a glass prism can be used to totally internally reflect a ray of light. (4 marks) A glass prism can be used to totally internally reflect a ray of light by ensuring the conditions for total internal reflection are met. Explanation: 1. Light must travel from a denser medium (glass) to a rarer medium (air). 2. The angle of incidence at the interface between the glass and air must be greater than the critical angle for that interface. A common application is using a 45^-45^-90^ prism. When a light ray enters one of the shorter faces normally, it travels straight to the hypotenuse. At the hypotenuse, the angle of incidence is 45^. Since the critical angle for a glass-air interface is typically around 41.8^, the 45^ angle of incidence is greater than the critical angle. This causes the light ray to undergo total internal reflection, reflecting off the hypotenuse back into the glass. The reflected ray then exits the prism, having been totally internally reflected. Diagram: (This diagram is conceptually similar to the one for Q21, showing light entering, reflecting off the hypotenuse, and exiting.) ` /|\ / | \ / | \ / | \ / | \ / | \ +------+------+ | | | | | | | | | | | | | | | | | | +------+------+ \ | / \ | / \ | / \ | / \ | / \|/ V ` (A right-angled isosceles prism. Light enters horizontally from the left, hits the hypotenuse, reflects downwards, and exits vertically. The angle of incidence at the hypotenuse is 45^, leading to total internal reflection.) 23. Explain two practical applications of total internal reflection. (4 marks) (i) Optical Fibers: Optical fibers use total internal reflection to transmit light signals over long distances with minimal loss. Light entering the fiber core (denser medium) strikes the cladding (rarer medium) at an angle greater than the critical angle, causing it to reflect repeatedly along the fiber. This is crucial for telecommunications and medical endoscopes. (ii) Periscopes and Binoculars: Prisms in periscopes and binoculars utilize total internal reflection to redirect light rays, allowing for compact designs and brighter images compared to using mirrors, which can absorb some light. For example, a pair of 45^-45^-90^ prisms can be arranged to achieve a 180^ turn in a periscope. 24. Define the following terms, giving their SI units: a) Work (1 mark) Work is done when a force causes a displacement of an object in the direction of the force. SI unit: Joule (J). b) Power (1 mark) Power is the rate at which work is done or energy is transferred. SI unit: Watt (W). c) Energy (1 mark) Energy is the capacity of a system to do work. SI unit: Joule (J). 25. A force of 200 N is applied to move a box horizontally through 10 m. Calculate the work done by the force. (3 marks) Step 1: Identify the given values. Force, F = 200 N Distance, d = 10 m The force is applied horizontally, and the box moves horizontally, so the force and displacement are in the same direction. Step 2: Apply the formula for work done. Work done (W) is calculated as the product of force and displacement in the direction of the force: W = F × d Step 3: Substitute the values and calculate the work done. W = 200 N × 10 m W = 2000 J The work done by the force is 2000 J. 3 done, 2 left today. You're making progress.