1. A point charge Q of mass 8.50 g hangs from the horizontal ceiling by a light 25.0-cm thread. When a horizontal electric field of magnitude 1750 N/C is turned on, the charge hangs away from the vertical as shown in the figure. The magnitude of Q is closest to:

Answer

Find vertical forces on the charge and solve for tension.

T\cos \theta=mg

T=\frac{mg}{\cos \theta}

Write an expression for the horizontal forces acting on the charge and solve for q.

T \sin \theta =qE

q=\frac{T \sin \theta}{E}=\frac{mg \sin \theta}{E \cos \theta}=\frac{mg \tan \theta}{E}=27.5\ \mu \textup{C}

2. A pair of charged conducting plates produces a uniform field of 12,000 N/C, directed to the right, between the plates. The separation of the plates is 40 mm. In the figure, an electron is projected from plate A, directly toward plate B, with an initial velocity of v_0=2.0\times10^7 \textup{m/s} . The velocity of the electron as it strikes plate B is closest to:

Answer

The change in kinetic energy equals the work done by the charge against the electric field.

W=\Delta KE

q_0Ed=\frac{1}{2}m_ev_f^2-\frac{1}{2}m_ev_0^2

Solving for v_f results in

v_f=\sqrt{\frac{m_ev_0^2+2q_eEd}{m_e}}=1.5\times10^7\ \textup{m/s}

3. Each of two small non-conducting spheres is charged positively, the combined charge being 50\ \mu \textup{C} . When the two spheres are 70\ \textup{cm} apart, each sphere is repelled from the other by a force of magnitude 4.0\ \textup{N} . Determine the magnitude of the smaller of the two charges.

Answer

The sum of both charges is

q_1+q_2=50\ \mu \textup{C} .

The force experienced by each charge is given by

F=k \frac{q_1q_2}{r^2}

Combining these two equations results in

F=k\frac{(q_{\textup{T}}-q_1)q_1}{r^2} .

Rearranging the above expression results in a quadratic equation.

q_1^2-q_{\textup T}q_1+\frac{Fr^2}{k}=0

Solving the quadratic equation results in q_1=4.8\ \mu \textup{C}

4. A glass sphere (n=1.6) is centered at the origin of a coordinate system, with its equatorial plane defining the xy-plane. A beam of light enters the glass sphere at a latitude of 40^{\circ} , parallel to the x-axis in the xz-plane. Make a careful drawing to determine the angle at which the beam will strike the back of the sphere. Will there be total internal reflection.

Answer

We first draw a detailed drawing.

Using Snell’s Law, we find \theta_2 as it enters the spherical glass.

n_1\sin \theta_1 = n_2 \sin \theta_2

\theta _2 =\sin^{-1} \left ( \frac{n_1}{n_2}\sin \theta_1 \right )=23.69^{\circ}

Next, we solve for \theta _3 .

\theta _3 = 180^{\circ}-2\theta_2=132.62^{\circ}

Finally, we find the angle at which the refracted ray hits the back of the spherical glass.

\theta_{\textup{refraction}}=180^{\circ}-40^{\circ}-132.62^{\circ} = 7.38^{\circ}

To find whether there will be total internal reflection, we must first find the critical angle for the glass-air interface.

\sin \theta_c=\frac{n_2}{n_1}

\theta _c=\sin ^{-1} \left ( \frac{1}{1.6} \right )=38.68^{\circ}

Since \theta _2 < \theta _c , there is no total internal reflection.

5. A thick glass plate (n=1.53) lies on the bottom of a tank of water (n=1.33). A light ray enters the water from air, making an angle of 72^{\circ} with the normal to the surface. What angle does the ray make with the normal when the ray is in the water? What angle does it make with the normal when it is in the glass?

Answer

We first draw a sketch of the situation to aid us in solving this problem.

Snell’s law is used to determine the angle \theta _w the ray makes with the normal when the ray is in the water.

n_a \sin \theta_a=n_w \sin \theta_w

\theta_w=\sin ^{-1} \left ( \frac{n_a \sin \theta_a}{n_w} \right )=45.65^{\circ}

Now that we know \theta _w , we can use Snell’s Law to find \theta _g .

n_w \sin \theta_w = n_g \sin \theta_g

\theta _g =\sin ^{-1} \left ( \frac{n_w \sin \theta _w}{n_g} \right ) = 38.43^{\circ}

6. A very wide light beam strikes a white screen at 90 degrees to the surface of the screen. An isosceles prism is placed in the way of the beam, as shown in the figure. How will the screen be illuminated if the index of refraction of the glass of the prism is n=1.5?

Answer

We start by drawing a detailed drawing of the situation.

We find \theta _p using Snell’s Law.

n_a \sin \theta_a = n_p \sin \theta_p

\theta_p = \sin ^{-1} \left ( \frac{n_a}{n_p} \sin \theta_a \right ) = 19.47 ^{\circ}

Next, we find \theta _{p2} by trigonometry.

\theta _4 = 90^{\circ} - \theta _p = 70.53^{\circ}

\theta_5 = 180^{\circ}-\theta_2 = 79.47^{\circ}

\theta_{p2} = 90^{\circ} - \theta_5 = 10.53^{\circ}

We find the angle of refraction \theta _{a2} by using Snell’s Law.

n_p \sin \theta_{p2} = n_a \sin \theta _{a2}

\theta_{a2}=\sin ^{-1} \left ( \frac{n_p}{n_a} \sin \theta_{p2} \right ) = 15.91^{\circ}

We find the value of y_2 :

x=\frac{1.5}{\tan 60^{\circ}} = 0.866\ \textup{cm}

y_2 =x \tan 10.53^{\circ} = 0.161\ \textup{cm}

Finally, we find y_1 :

y_1 = x_2 \tan \theta _{a2} = 0.57\ \textup{cm}

With this information, we are able to draw a detailed sketch of where the rays appear behind the triangle.

7. Consider light that is perpendicularly incident on a triangular prism of the kind shown in the figure. The index of refraction of the prism material is n_1=1.814 . Suppose that two reflecting sides are coated with a thin, uniform layer of a dielectric with index of refraction n_2=1.380 . Will the glass-dielectric interface be totally reflecting? How large can n_2 be so that the interface is still totally reflecting?

Answer

We find the critical angle for the configuration above.

\sin \theta_c=\frac{n_2}{n_1}

\theta_c = \sin ^{-1} \frac{1.380}{1.814}

\theta_c = 49.53^{\circ}

In order for the prism to be totally reflecting, the incidence angle must be greater than the critical angle. Since the angle of incidence is 45^{\circ} , the prism is not totally reflecting.

Next, we find how large n_2 can be so that the interface is still totally reflecting.

For the prism to be totally reflecting, the following must hold: \theta_2 \ge \theta_c.

Setting the critical angle to 45^{\circ} , we find the cutoff value for n_2 .

n_{2c}=n_1\sin 45^{\circ}=1.28

It follows that n_2\le1.28 for total internal reflection.

8. Light is incident on an equilateral triangular prism (n=1.55) at a 35^{\circ} from the normal to the normal to one of the faces. What is the exit angle?

Answer

We redraw the diagram and include the variables we want to solve for.

\theta_1 can be found by using Snell’s Law.

\theta_1 = sin^{-1} \left ( \frac{n_a\sin\theta_a}{n_p} \right ) = 21.72 ^{\circ}

Given that we’re dealing with an equilateral triangle, and the sum of the interior angles is 180^{\circ} , we can find \theta_2 and \theta_4 and \theta_5

\theta_2 = 90^{\circ}-21.72^{\circ}=68.28^{\circ}

\theta_4 = 180^{\circ}-60^{\circ}-68.28^{\circ}=51.72^{\circ}

\theta_5 = 90^{\circ}-51.72^{\circ}=38.28^{\circ}

Finally, we use Snell’s Law to find the exit angle.

\theta_E=\sin^{-1} \left ( \frac{n_p\sin \theta_3}{n_a} \right )=73.79^{\circ}

9. A dime 60 cm from and on the optical axis of a concave spherical mirror produces an image 20 cm away from the mirror. If the dime is moved on the axis to 35 cm from the mirror, where will the image move? How large is the radius of the sphere of which the mirror is a section? Draw the system for the second case described.

For this problem, we can determine the focal length by using the source-image-focal-length relation.

\frac{1}{s}+\frac{1}{i}=\frac{1}{f}

\frac{1}{\textup {60 cm}}+\frac{1}{\textup {20 cm}}=\frac{1}{f}

f= \textup{15 cm}

Next, we find the distance i for when the object is moved 35 cm from the mirror.

\frac{1}{s}+\frac{1}{i}=\frac{1}{f}

\frac{1}{\textup {35 cm}} +\frac{1}{i}=\frac{1}{\textup{15 cm}}

i=\textup{26.25 cm}

Finally, we find the radius of the sphere.

R=2f=\textup{30 cm}

10. At noon, a 2.0-m-long vertical stick casts a shadow 1.0 m long. If the same stick is placed in a flat-bottomed pool of water half the height of the stick (still at noon), how long is the shadow on the floor of the pool? (For water, n=1.33.)

We solve for the angle.

\theta = \tan ^{-1} \left ( \frac{H}{x} \right)=\tan ^{-1} \left ( \frac{\textup{2 m}}{\textup{1 m}} \right)=63.43^{\circ}

We draw the situation for when the vertical stick is placed in the pool.

We find \theta_2 using Snell’s Law.

\theta_1=90^{\circ}-63.43^{\circ}=26.57^{\circ}

n_1\sin \theta_1 = n_2 \sin \theta_2

\theta_2 = \sin ^{-1}\left ( \frac{n_1}{n_2}\sin \theta_1 \right ) = 19.65^{\circ}

We now find x_1 .

x_1=\frac{H}{2\tan \theta} = \textup {0.5 m}

We find x_2 .

x_2 = \frac{H\tan \theta_3}{2}=\textup {0.3571 m}

The length of the shadow is

x_1+x_2=\textup {0.8571 m} .

11. A concave mirror is cut from a spherical surface of radius of curvature 2.0 m. A pencil 10 cm long is placed perpendicular to the axis of the mirror at a distance of 80 cm from the mirror. Where is the image and how large is it?

We find the focal length f .

f=\frac{R}{2}=\textup{1 m} .

We then find the image distance i .

\frac{1}{s}+\frac{1}{i}=\frac{1}{f}

i=\textup{-4 m}

The fact that the image distance is negative means that it’s a virtual image.

To find the height of the image, we use the magnification equation:

M=-\frac{i}{s}=\frac{h_i}{h_s}

h_i=-\frac{ih_s}{s}=\textup{0.5 m}