(I) The critical angle for a certain liquid-air surface is 44.7º. What is the index of refraction of the liquid?

Since the **critical angle** is 44.7º, we can set up the equation_{1} and solving for n_{2} gives n_{2} = 1.422.

sin(44.7º) = nPlugging in 1.0003 for n_{1}/ n_{2}

(II) A ray of light enters a light fiber at an angle of 15º with the long axis of the fiber, as in Fig. 23-54. Calculate the distance the light ray travels between successive reflections off the sides of the fiber. Assume that the fiber has an index of refraction of 1.6 and is 10^{-4}m in diameter.

The answer can be found using trigonometry. You can set up a right triangle, with one leg being the diameter of the fiber, and the hypoteneuse being the distance that the light actually travels. The angle is 15º, so sin(15º) = opposite / hypoteneuse, which is the diameter over the distance light travels. Thus, our equation is^{-4}m.

sin(15º) = 10Solving for x gives us x = 3.86 x 10^{-4}m / x

(II) A stamp collector uses a converging lens with focal length 24 cm to view a stamp 18 cm in front of the lens. (*a*) Where is the image located? (*b*) What is the magnification?

The answer to all of these lies in the equation_{o} (we can ignore the conversion from cm to m because it would cancel out through every term anyways), we get d_{i} = -72 cm.

Finding the magnification is a simple process, as m = -d_{i} /d_{o}. Plugging in -72 and 18 gives us m = -(-72)/(18) = 4.

1/f = 1/dFor part a, plugging in 24 for f and 18 for d_{o}+ 1/d_{i}

Finding the magnification is a simple process, as m = -d

(II) An 80-mm-focal-length lens is used to focus an image on the film of a camera. The maximum distance allowed between the lens and the film plane is 120 mm. (*a*) how far ahead of the film should the lens be if the object to be photographed is 10.0 m away? (*b*) 3.0 m away? (*c*) 1.0 m away? (*d*) What is the closest object this lens could photograph sharply?

Once again, all we need is the equation

Part a has a d_{o} of 10, and plugging in gives a d_{i} of -0.0806 m.

Part b has a d_{o} of 3, which gives a d_{i} of 0.082 m.

Part c has a d_{o} of 1, which gives a d_{i} of 0.0869 m.

Part d is slightly different: we\'re solving for d_{o} given d_{i}. As the problem states, \"The maximum distance allowed between the lens and the film plane is 120 mm.\" Therefore, in this case, d_{i} is 0.12 m. Plugging that in gives us a d_{o} of 0.24 m.

1/f = 1/dThe f throughout is 0.08 m (I have converted to meters for convenience)._{o}+ 1/d_{i}

Part a has a d

Part b has a d

Part c has a d

Part d is slightly different: we\'re solving for d

(II) (*a*) An object 31.5 cm in front of a certain lens is imaged 8.20 cm in front of that lens (on the same side as the object). What type of lens is this and what is its focal length? Is the image real or virtual? (*b*) If the image were located, instead, 38.0 cm in front of the lens, what type of lens would it be and what focal length would it have?

From the equation 1/f = 1/d_{o} + 1/d_{i}, we can find the focus. d_{o} is 31.5 cm (again, we\'re not converting for convenience), and d_{i} is -8.2 (because it\'s on the same side). Therefore, f = -11.086 cm. We know that the lens is diverging because f is negative. We also know that the lens is concave, because it\'s diverging. And the image is virtual because d_{i} is negative, meaning it\'s on the same side of the lens and virtual.

If the image were located 38 cm in front of the lens, then we get a different value for f. We get f = 184.154 cm, which means that the lens is a converging lens (since f is positive). Since it\'s converging, the lens must be concave. And, since d_{i} is negative again, the image is virtual again.

If the image were located 38 cm in front of the lens, then we get a different value for f. We get f = 184.154 cm, which means that the lens is a converging lens (since f is positive). Since it\'s converging, the lens must be concave. And, since d