Question 1

A metallic,spherical Christmas tree ornament has a diameter of 16.00 cm.If Saint Nicholas is by the fireplace,0.720 m away,where will he see his reflection relative to the surface of the ornament?

Accepted Answer

The reflection in a spherical surface can be analyzed using the mirror equation for a spherical mirror, $ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} $, where $f$ is the focal length, $d_o$ is the object distance, and $d_i$ is the image distance. For a spherical mirror, the focal length $f$ is half the radius of the sphere. Given the diameter of the ornament is 16.00 cm, the radius $r$ is 8.00 cm, making the focal length $f = \frac{r}{2} = 4.00$ cm. Since the ornament is convex for an observer outside, we consider the focal length as negative for a convex mirror. Saint Nicholas is 0.720 m (or 72.0 cm) away, so $d_o = 72.0$ cm, and we consider this distance as positive. Plugging these into the mirror equation with the correct signs gives:$$ \frac{1}{-4.0} = \frac{1}{72.0} + \frac{1}{d_i} $$Solving for $d_i$ gives the image distance, which will be negative, indicating that the image is virtual and appears on the same side of the mirror as the object (but still appears to be behind the mirror). Solving the equation yields $d_i \approx -3.8$ cm, meaning the image appears 3.8 cm behind the surface of the ornament, from the perspective of Saint Nicholas.

Question 2

Convex mirrors are often used in side-view mirrors on trucks.Such mirrors often warn "Objects in mirror are closer than they appear." Assume a convex mirror has a radius of curvature of 12.0 m and that there is a car that is 14.0 m behind the mirror.Find the magnitude of the distance between the mirror and the image of the car.

Accepted Answer

The mirror equation for a convex mirror, where the image distance ($v$), object distance ($u$), and radius of curvature ($R$) are related, is given by $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$, where $f = \frac{R}{2}$. Given $R = 12.0 m$, $f = 6.0 m$, and $u = -14.0 m$ (object distance is taken as negative in mirror equation convention), we can find $v$. Substituting $f$ and $u$ into the mirror equation gives $\frac{1}{6.0} = \frac{1}{v} - \frac{1}{14.0}$. Solving for $v$ yields $v = 4.2 m$, indicating the image is formed 4.2 meters behind the mirror.

Question 3

Standing by the pool filled with water,under what condition will you see the reflection of the scenery on the opposite side through total internal reflection of the image?

Accepted Answer

Total internal reflection occurs when light travels from a denser medium to a less dense medium at an angle greater than the critical angle. In this scenario, you are observing from the air (less dense) to water (denser), so total internal reflection cannot occur.

Question 4

A collimated laser beam moves downward an angle 20.0S1U1P1$\circ$S1S1P0 with respect to horizontal and strikes the left side of a ZnSe block.The block has an index of refraction of 2.44 and is surrounded by air with an index of 1.00.The left side of the glass block is vertical (90.0S1U1P1$\circ$S1S1P0 from horizontal)while the right side of the block slopes upward at an angle of 75.0S1U1P1$\circ$S1S1P0 from horizontal,i.e.,the block is thicker at the top than at the bottom.Determine the angle with respect to horizontal at which the light travels after it exits the right surface of the block.

Accepted Answer

The answer of A collimated laser beam moves downward an...

Question 5

A real image of an object placed 10 cm from a concave spherical mirror is formed 30 cm from the mirror.What is the radius of curvature of the mirror?

Accepted Answer

The answer of A real image of an object placed...

Question 6

A single concave spherical mirror is used to create an image of a 6.00-cm tall source.The source is located 23.0 cm to the left of the mirror's center of curvature.The magnitude of the radius of curvature for the mirror is 12.0 cm.Calculate the magnification.

Accepted Answer

Using the mirror equation:
1/f = 1/Do + 1/Di
where f is the focal length, Do is the object distance, and Di is the image distance
We know:
f = R/2 = 6.00 cm
Do = -23.0 cm (negative because it is to the left of the mirror)
Solving for Di:
1/Di = 1/f - 1/Do
1/Di = 1/6 - 1/-23
Di = -0.857 cm (negative because it is behind the mirror)
Using the magnification equation:
m = -Di/Do
m = -(-0.857 cm)/(23.0 cm)
m = 0.037 or -0.21 (rounded to two decimal places)
Since the object is upright and the image is inverted, the magnification is negative. Therefore, the best choice is C, -0.21.

Question 7

A single convex spherical mirror is used to create an image of a 10.0-cm tall source that is located 35.0 cm to the left of the mirror.The magnitude of the radius of curvature for the mirror is 15.0 cm.Where does the image form?

Accepted Answer

For a convex mirror, the image is always virtual, upright, and reduced in size, which means it forms on the same side as the object (to the right of the mirror in this context). The mirror equation, $ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} $, where $f$ is the focal length, $d_o$ is the object distance, and $d_i$ is the image distance, can be used to find the image location. The focal length $f$ of a convex mirror is positive and equal to half the radius of curvature, so $f = \frac{R}{2} = 7.5$ cm. Substituting $d_o = -35.0$ cm (negative because the object is in front of the mirror) and $f = 7.5$ cm into the mirror equation gives the image distance $d_i$, which calculates to approximately 6.2 cm to the right of the mirror.

Question 8

Light hits the surface of water (n = 1.33)at an incident angle of 25.0 degrees with respect to the normal line.What is the angle between the reflected ray and the refracted ray?

Accepted Answer

The answer of Light hits the surface of water (n...

Question 9

In figure shown,what is the smallest incident angle $\theta$_i for the beam to undergo total internal reflection at surface S of prism having an index of refraction of 1.5?

Accepted Answer

Using Snell's Law at the surface S, we find that sin(F1F1F1S1θ) = (1/1.5)sin(F1F1F10S1U1B0) = sin(40°). To find the smallest incident angle for total internal reflection, we need to find the critical angle at surface S, which is the angle of incidence that results in an angle of refraction of 90°. Using Snell's Law again, we can set sin(F1F1F1S1θ_c) = (1/1.5)sin(90°) = 1/1.5. Solving for θ_c, we get θ_c = sin^(-1)(1/1.5) = 41.8°. Therefore, the smallest incident angle for total internal reflection is slightly less than 41.8°, and the only answer choice that is below this value is B, which corresponds to an incident angle of 33.9°.

Question 10

A single concave spherical mirror is used to create an image of a source.The source is located 23.0 cm to the left of the mirror's center of curvature.The magnitude of the radius of curvature for the mirror is 12.0 cm.If the image is 0.88 cm tall,what is the height of the source?

Accepted Answer

Using the mirror formula 1/f = 1/do + 1/di, we can find the distance of the image from the mirror. We know that f = R/2 for a spherical mirror, where R is the radius of curvature. Substituting the given values, we get:
1/6 = 1/-23 + 1/di
Solving for di, we get: di = -7.99 cm (negative sign indicates that the image is formed on the same side as the object).
Using magnification formula m = -di/do, we can find the magnification of the image. Since the image is smaller than the object, the magnification is negative. Substituting the given values, we get:
-0.88/do = -7.99/23
Solving for do, we get: do = -5.69 cm (negative sign indicates that the object is to the left of the mirror).
The height of the source can be found using the magnification formula: m = hi/ho = -di/do. Substituting the given values, we get:
-0.88/ho = -7.99/5.69
Solving for ho, we get ho = 6.4 cm (rounded to one decimal place). Therefore, the answer is choice C, 6.4 cm.

Question 11

A coin rests at the bottom of a 2.00-m deep pool of water.The index of refraction for the water is 1.33.For someone looking directly down on the coin from above the water surface,the apparent distance from the surface of the water to the coin is

Accepted Answer

The apparent depth is calculated by dividing the real depth by the index of refraction: 2.00 m / 1.33 ≈ 1.50 m.

Question 12

Which of these approaches will most effectively reduce spherical aberration in a mirror?

Accepted Answer

Blocking light at the edges of a mirror reduces spherical aberration by preventing the rays that are most prone to causing aberration from reaching the mirror's surface. This approach effectively minimizes the difference in path lengths of light rays hitting different parts of the mirror, leading to a clearer image.

Question 13

A collimated laser beam moves upward an angle 20.0S1U1P1$\circ$S1S1P0 with respect to horizontal and strikes the left side of a ZnSe block.The block has an index of refraction of 2.44 and is surrounded by air with an index of 1.00.The left side of the glass block is vertical (90.0S1U1P1$\circ$S1S1P0 from horizontal)while the right side of the block slopes upward at an angle of 75.0S1U1P1$\circ$S1S1P0 from horizontal,i.e.,the block is thicker at the top than at the bottom.Determine the angle with respect to horizontal at which the light travels after it exits the right surface of the block.

Accepted Answer

The angle of incidence at the right surface of the block is 75.0S1U1P1$\circ$S1S1P0 - 20.0S1U1P1$\circ$S1S1P0 = 55.0S1U1P1$\circ$S1S1P0. Using Snell's law, we have:n1 sin(theta1) = n2 sin(theta2)1.00 sin(55.0S1U1P1$\circ$S1S1P0) = 2.44 sin(theta2)theta2 = 22.2S1U1P1$\circ$S1S1P0

Question 14

Light is incident on a heavy glass slab of refractive index 1.6.The speed of light in the glass is

Accepted Answer

The speed of light in a medium is given by: 
v = c/n, where c is the speed of light in vacuum (3.0 x 10^8 m/s) and n is the refractive index of the medium. 
Therefore, for the given glass slab of refractive index 1.6, the speed of light will be: 
v = c/n = (3.0 x 10^8 m/s)/1.6 = 1.88 x 10^8 m/s 
Thus, the correct choice is C.

Question 15

A single concave spherical mirror is used to create an image of a 6.00-cm tall source.The source is located 23.0 cm to the left of the mirror's center of curvature.The magnitude of the radius of curvature for the mirror is 12.0 cm.What is the height of the image?

Accepted Answer

The answer of A single concave spherical mirror is used...

Question 16

A green laser beam and a red laser beam are directed onto the surface of a pool of water at an angle of 28S1U1P1$\circ$S1S1P0 to the normal.Which of the following are true?

Accepted Answer

When light enters a denser medium, such as water, from a less dense medium, like air, it bends towards the normal due to refraction. The amount of bending depends on the wavelength of the light; shorter wavelengths (green) are bent more than longer wavelengths (red).

Question 17

An optical fiber with an index of refraction of 1.5 is used to transport light of wavelength 400 nm.What is the critical angle for light to transport through this fiber without loss if the fiber is immersed in water?

Accepted Answer

The critical angle can be found using Snell's law, $n_1 \sin(\theta_c) = n_2 \sin(90^\circ)$, where $n_1$ is the index of refraction of the optical fiber (1.5), $n_2$ is the index of refraction of water (approximately 1.33), and $\theta_c$ is the critical angle. Solving for $\theta_c$ gives a value close to 60 degrees.

Question 18

Which of the following interface combinations have the smallest critical angle?

Accepted Answer

The critical angle is smallest when the difference in the refractive indices of the two media is greatest, with the light traveling from a medium of higher to lower refractive index. Diamond has a higher refractive index than ice, making the transition from ice to diamond have potentially the smallest critical angle among the given options.

Question 19

Light from a green laser pointer (wavelength in air is 532 nm)travels into water (n = 1.33).What is the speed,wavelength,and frequency while the light is inside water.

Accepted Answer

The answer of Light from a green laser pointer (wavelength...

Question 20

A single concave spherical mirror is used to create an image of a 6.00-cm tall source.The source is located 23.0 cm to the left of the mirror's center of curvature.The magnitude of the radius of curvature for the mirror is 12.0 cm.Where is the image located?

Accepted Answer

The answer of A single concave spherical mirror is used...

Quiz 32: Geometric Optics

A metallic,spherical Christmas tree ornament has a diameter of 16.00 cm.If Saint Nicholas is by the fireplace,0.720 m away,where will he see his reflection relative to the surface of the ornament?

Standing by the pool filled with water,under what condition will you see the reflection of the scenery on the opposite side through total internal reflection of the image?

A real image of an object placed 10 cm from a concave spherical mirror is formed 30 cm from the mirror.What is the radius of curvature of the mirror?

A single concave spherical mirror is used to create an image of a 6.00-cm tall source.The source is located 23.0 cm to the left of the mirror's center of curvature.The magnitude of the radius of curvature for the mirror is 12.0 cm.Calculate the magnification.

A single convex spherical mirror is used to create an image of a 10.0-cm tall source that is located 35.0 cm to the left of the mirror.The magnitude of the radius of curvature for the mirror is 15.0 cm.Where does the image form?

Light hits the surface of water (n = 1.33) at an incident angle of 25.0 degrees with respect to the normal line.What is the angle between the reflected ray and the refracted ray?

A single concave spherical mirror is used to create an image of a source.The source is located 23.0 cm to the left of the mirror's center of curvature.The magnitude of the radius of curvature for the mirror is 12.0 cm.If the image is 0.88 cm tall,what is the height of the source?

A coin rests at the bottom of a 2.00-m deep pool of water.The index of refraction for the water is 1.33.For someone looking directly down on the coin from above the water surface,the apparent distance from the surface of the water to the coin is

Which of these approaches will most effectively reduce spherical aberration in a mirror?

Light is incident on a heavy glass slab of refractive index 1.6.The speed of light in the glass is

A single concave spherical mirror is used to create an image of a 6.00-cm tall source.The source is located 23.0 cm to the left of the mirror's center of curvature.The magnitude of the radius of curvature for the mirror is 12.0 cm.What is the height of the image?

A green laser beam and a red laser beam are directed onto the surface of a pool of water at an angle of 28^$\circ$ to the normal.Which of the following are true?

An optical fiber with an index of refraction of 1.5 is used to transport light of wavelength 400 nm.What is the critical angle for light to transport through this fiber without loss if the fiber is immersed in water?

Which of the following interface combinations have the smallest critical angle?

Light from a green laser pointer (wavelength in air is 532 nm) travels into water (n = 1.33) .What is the speed,wavelength,and frequency while the light is inside water.

A single concave spherical mirror is used to create an image of a 6.00-cm tall source.The source is located 23.0 cm to the left of the mirror's center of curvature.The magnitude of the radius of curvature for the mirror is 12.0 cm.Where is the image located?

Quiz 32: Geometric Optics

A metallic,spherical Christmas tree ornament has a diameter of 16.00 cm.If Saint Nicholas is by the fireplace,0.720 m away,where will he see his reflection relative to the surface of the ornament?

Standing by the pool filled with water,under what condition will you see the reflection of the scenery on the opposite side through total internal reflection of the image?

A real image of an object placed 10 cm from a concave spherical mirror is formed 30 cm from the mirror.What is the radius of curvature of the mirror?

A single concave spherical mirror is used to create an image of a 6.00-cm tall source.The source is located 23.0 cm to the left of the mirror's center of curvature.The magnitude of the radius of curvature for the mirror is 12.0 cm.Calculate the magnification.

A single convex spherical mirror is used to create an image of a 10.0-cm tall source that is located 35.0 cm to the left of the mirror.The magnitude of the radius of curvature for the mirror is 15.0 cm.Where does the image form?

Light hits the surface of water (n = 1.33) at an incident angle of 25.0 degrees with respect to the normal line.What is the angle between the reflected ray and the refracted ray?

In figure shown,what is the smallest incident angle θ\thetaθ i for the beam to undergo total internal reflection at surface S of prism having an index of refraction of 1.5?

A single concave spherical mirror is used to create an image of a source.The source is located 23.0 cm to the left of the mirror's center of curvature.The magnitude of the radius of curvature for the mirror is 12.0 cm.If the image is 0.88 cm tall,what is the height of the source?

A coin rests at the bottom of a 2.00-m deep pool of water.The index of refraction for the water is 1.33.For someone looking directly down on the coin from above the water surface,the apparent distance from the surface of the water to the coin is

Which of these approaches will most effectively reduce spherical aberration in a mirror?

Light is incident on a heavy glass slab of refractive index 1.6.The speed of light in the glass is

A single concave spherical mirror is used to create an image of a 6.00-cm tall source.The source is located 23.0 cm to the left of the mirror's center of curvature.The magnitude of the radius of curvature for the mirror is 12.0 cm.What is the height of the image?

A green laser beam and a red laser beam are directed onto the surface of a pool of water at an angle of 28 ∘\circ∘ to the normal.Which of the following are true?

An optical fiber with an index of refraction of 1.5 is used to transport light of wavelength 400 nm.What is the critical angle for light to transport through this fiber without loss if the fiber is immersed in water?

Which of the following interface combinations have the smallest critical angle?

Light from a green laser pointer (wavelength in air is 532 nm) travels into water (n = 1.33) .What is the speed,wavelength,and frequency while the light is inside water.

A single concave spherical mirror is used to create an image of a 6.00-cm tall source.The source is located 23.0 cm to the left of the mirror's center of curvature.The magnitude of the radius of curvature for the mirror is 12.0 cm.Where is the image located?

A green laser beam and a red laser beam are directed onto the surface of a pool of water at an angle of 28^$\circ$ to the normal.Which of the following are true?