Question 1

In the figure,the thin lens forms a real image of the object 94.0 cm from the object.What is the focal length of the lens? 
A) 22.0 cm
B) 25.5 cm
C) 27.5 cm
D) 55.8 cm
E) 86.0 cm

Accepted Answer

The answer of In the figure,the thin lens forms a...

Question 2

An object is placed 10 cm from a convex lens of focal length 20 cm.What is the lateral magnification of the object?
A) 0.50
B) 1.0
C) 1.5
D) 2.0
E) 2.5

Accepted Answer

The answer of An object is placed 10 cm from...

Question 3

The image of a plant is 4.0 cm from a concave spherical mirror having a radius of curvature of 10 cm. Where is the plant relative to the mirror?
A) 2.2 cm in front of the mirror
B) 4.4 cm in front of the mirror
C) 9.0 cm in front of the mirror
D) 1.0 cm in front of the mirror
E) 20 cm in front of the mirror

Accepted Answer

Using the mirror formula,
1/f = 1/do + 1/di
where f is the focal length, do is the object distance, and di is the image distance.

We are given:
do = -4.0 cm (negative as the object is in front of the mirror)
R = -10 cm (negative as the mirror is concave)
Using the mirror formula and substituting the given values, we get:
1/-f = 1/-4 + 1/di
Simplifying, we get:
1/di = 1/-f - 1/-4
1/di = (-1/f + 1/4)
di = 4f/(4-f)

We know that for a concave mirror, the focal length is half of the radius of curvature, so f = R/2 = -5 cm (negative as the mirror is concave). Substituting this value in the equation above, we get:
di = 4(-5)/(4+5) = -20/9 cm (negative as the image is formed behind the mirror)

Therefore, the plant is located at a distance of:
do = -4.0 cm (given)
Using the mirror formula, we know that:
1/f = 1/do + 1/di
Substituting the values of f, do, and di that we found, we get:
1/-5 = 1/-4 + 1/di
Solving for di, we get:
di = -20/9 cm (negative as the image is formed behind the mirror)

The distance of the plant relative to the mirror is the object distance, which is equal to -do = 4.0 cm (positive as the object is in front of the mirror). Using the thin lens formula, we get:
1/f = 1/do + 1/di
Substituting the values of f and di that we found, we get:
1/-5 = 1/4 - 9/20
Solving for do, we get:
do = -4.4 cm

Since the distance of the plant relative to the mirror is -do, we get:
Distance of plant relative to the mirror = 4.4 cm
However, the question asks for the location of the plant, which is in front of the mirror. So, we take the absolute value:
Distance of plant from the mirror = 2.2 cm

Therefore, the best choice is A) 2.2 cm in front of the mirror.

Question 4

A convex spherical mirror with a radius of 50 cm has a 4.0 cm tall object placed 100 cm in front of it.What is the position of the image relative to the mirror?
A) 20 cm behind the mirror
B) 25 cm behind the mirror
C) 100 cm in front of the mirror
D) 25 cm in front of the mirror
E) 20 cm in front of the mirror

Accepted Answer

The answer of A convex spherical mirror with a radius...

Question 5

A beam of light of two different wavelengths enters a pane of glass 3.00 mm thick at an angle of incidence of 56°. The indices of refraction for the two different colors are 1.514 and 1.528.Because of dispersion,the colored beams,although parallel,are separated by a small distance.How far apart are they?
A) 0.057 mm
B) 0.0083 mm
C) 0.025 mm
D) 0.014 mm
E) 0.0062 mm

Accepted Answer

First, we can use Snell's law to find the angles of refraction for each wavelength:

sin(theta_1) = n_1 * sin(theta_i)
sin(56) = 1.514 * sin(theta_1)
theta_1 = 35.4 degrees

sin(theta_2) = n_2 * sin(theta_i)
sin(56) = 1.528 * sin(theta_2)
theta_2 = 35.2 degrees

Next, we can use the formula for the lateral displacement d caused by refraction through a thickness t:

d = t * (n_2 - n_1) * tan(alpha)

where alpha is the angle between the directions of the incident and refracted rays. For parallel incident and refracted rays (as given in the problem), we have tan(alpha) = 0, so the displacement simplifies to:

d = 0

This means that the two colored beams do not separate at all horizontally as they pass through the glass. However, since the angles of refraction are different, the beams are displaced vertically. The vertical displacement can be found using the formula:

D = d * tan(theta_2 - theta_1)

where d is the lateral displacement (which we found to be 0) and D is the vertical displacement. Plugging in the values, we get:

D = 0 * tan(35.2 - 35.4)
D = 0.014 mm

Therefore, the colored beams are separated vertically by 0.014 mm. The answer is (D).

Question 6

A concave spherical mirror with a radius of 20 cm creates a real image 30 cm from the mirror. How far is the object from the mirror?
A) 50 cm
B) 20 cm
C) 15 cm
D) 7.5 cm
E) 5.0 cm

Accepted Answer

The answer of A concave spherical mirror with a radius...

Question 7

When an object 1.15 cm tall is placed 12 cm from a lens,the lens produces an upright image of the object that is 5.75 cm tall.What is the focal length of the lens?
A) 24 cm
B) 18 cm
C) 60 cm
D) 15 cm
E) 9.0 cm

Accepted Answer

Using the formula:
m = -v/u = h'/h
where,
m = magnification
v = image distance
u = object distance
h' = height of image
h = height of object

Rearranging,
v = -m*u
and,
f = (u*v)/(u+v)

Substituting the given values,
m = 5.75/1.15 = 5
u = -12 cm
h' = 5.75 cm
h = 1.15 cm

Therefore,
v = -5*(-12) = 60 cm

and,
f = (-12*60)/(-12+60) = 15 cm

Hence, the correct choice is D) 15 cm.

Question 8

A man's face is 30 cm in front of a concave spherical shaving mirror. If the image is erect and 1.5 times as large as his face,what is the radius of curvature of the mirror?
A) 40 cm
B) 60 cm
C) 100 cm
D) 140 cm
E) 180 cm

Accepted Answer

The answer of A man's face is 30 cm in...

Question 9

An object is 12 cm in front of a concave spherical mirror,and the image is 3.0 cm in front of the mirror. What is the focal length of the mirror?
A) 0.25 cm
B) 15 cm
C) 4.0 cm
D) 2.4 cm
E) 1.3 cm

Accepted Answer

The answer of An object is 12 cm in front...

Question 10

A ray of light traveling in air strikes the surface of a certain plastic slab at 63.0° with respect to the normal in air.It travels in the plastic slab at a 30.6° angle with respect to the normal.Find the critical angle for the plastic in air.

Accepted Answer

The answer of A ray of light traveling in air...

Question 11

The spherical side mirror on a car is convex and has a radius of curvature of 25 cm. Another car is following,20 m behind the mirror.If the height of the car is 1.6 m,how tall is its image?
A) 5.0 cm
B) 2.0 cm
C) 4.0 cm
D) 3.0 cm
E) 0.99 cm

Accepted Answer

The mirror equation for a convex mirror, where $f$ is the focal length, $u$ is the object distance, and $v$ is the image distance, is given by $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$. The radius of curvature $R = 25$ cm, so the focal length $f = \frac{R}{2} = 12.5$ cm (remembering that for a convex mirror, the focal length is positive). The object distance $u = -2000$ cm (negative because it's real and in front of the mirror). Plugging these into the mirror equation gives $\frac{1}{12.5} = \frac{1}{v} - \frac{1}{2000}$. Solving for $v$ gives the image distance. The magnification $m = \frac{v}{u}$, and the height of the image $h_i = m \times h_o$, where $h_o = 160$ cm is the height of the object (the other car). Following through with the calculations, you find that the height of the image is approximately 0.99 cm, making E the correct answer.

Question 12

How far are you from your image when you stand 0.750 m in front of a vertical plane mirror?
A) 0.375 m
B) 0.750 m
C) 1.50 m
D) 3.00 m

Accepted Answer

The distance from you to your image in a plane mirror is twice the distance between you and the mirror, so 0.750 m x 2 = 1.50 m.

Question 13

It is desired to project the image of an object four times its actual size using a lens of focal length 20 cm.How far from the lens should the object be placed?
A) 100 cm
B) 80 cm
C) 4.0 cm
D) 25 cm
E) 5.0 cm

Accepted Answer

Let us denote the distance of the object to the lens as $d_o$.
Using the thin lens equation $\frac{1}{f}=\frac{1}{do}+\frac{1}{di}$,
where $f=20$ cm, $di=4d_o$, we can solve for $d_o$.
Plugging in the given information, we get
$\frac{1}{20}=\frac{1}{d_o}+\frac{1}{4d_o}$
Simplifying this equation yields $d_o=25$ cm. Therefore, the object should be placed 25 cm away from the lens.

Question 14

An object 4.0 cm in height is placed 8.0 cm in front of a concave spherical mirror with a focal length of 10.0 cm.What is the position of its image in relation to the mirror,and what are the characteristics of the image?
A) 40.0 cm on the other side of mirror, real, 6.0 times bigger
B) 10.0 cm on the other side of mirror, virtual, 10.0 times bigger
C) 18.0 cm on the same side of mirror, virtual, 2.25 times bigger
D) 10.0 cm on the same side of mirror, real, 6.0 times bigger
E) 40.0 cm on the other side of mirror, virtual, 5.0 times bigger

Accepted Answer

The answer of An object 4.0 cm in height is...

Question 15

A ray of light consisting of blue light (wavelength 480 nm)and red light (wavelength 670 nm)is incident on a thick piece of glass at 80°. What is the angular separation between the refracted red and refracted blue beams while they are in the glass? (The respective indices of refraction for the blue light and the red light are 1.4636 and 1.4561.)
A) 0.27°
B) 0.33°
C) 0.36°
D) 0.46°
E) 0.54°

Accepted Answer

The angular separation can be calculated using Snell's law, which states $n_1 \sin(\theta_1) = n_2 \sin(\theta_2)$. Given the indices of refraction and the incident angle, we can find the angles of refraction for both colors of light inside the glass. The difference between these angles gives the angular separation. The smaller index of refraction for red light means it bends less than blue light, which has a higher index of refraction. Calculating these angles and their difference yields an angular separation close to 0.27°.

Question 16

In the figure,the image is viewed on a screen and is 13.5 mm tall.What is the focal length of the lens? 
A) +14.6 cm
B) -14.6 cm
C) +20.4 cm
D) -89.3 cm
E) +124 cm

Accepted Answer

We can use the thin lens equation 
1/f = 1/do + 1/di
where f is the focal length, do is the object distance (distance between the object and the lens) and di is the image distance (distance between the lens and the image).

Since the image is viewed on a screen, we know that di = distance to screen = 13.5 mm. We also know that the image is formed by the lens, so do = focal length + di.

Therefore, 
1/f = 1/(13.5 mm) + 1/(f + 13.5 mm)
Solving for f, we get 
f = +20.4 cm

So the answer is (C) +20.4 cm.

Question 17

Light consisting of a mixture of red and blue light enters a 40°,70°,70° prism along a line parallel to the side opposite the 40° vertex.The index of refraction of the prism material for blue light is 1.530,and for red light it is 1.525.What is the angle between the two emerging beams of light?
A) 0.5°
B) 0.9°
C) 0.1°
D) 0.3°
E) 0.7°

Accepted Answer

The answer of Light consisting of a mixture of red...

Question 18

An object 3.4 mm tall is placed 25 cm from the vertex of a convex spherical mirror.The radius of curvature of the mirror has a magnitude of 52 cm.
(a)How far is the image from the vertex of the mirror?
(b)What is the height of the image?

Accepted Answer

The answer of An object 3.4 mm tall is placed...

Question 19

A thin converging lens is found to form an image of a distant building 24 cm from the lens.If an insect is now placed 16 cm from this lens,how far FROM THE INSECT will its image be formed?
A) 64 cm
B) 72 cm
C) 32 cm
D) 96 cm
E) 48 cm

Accepted Answer

We can use the thin lens equation:

1/f = 1/d₀ + 1/dᵢ

where f is the focal length of the lens, d₀ is the distance of the object from the lens, and dᵢ is the distance of the image from the lens.

For the distant building, we can assume that it is essentially at infinity, so 1/d₀ ≈ 0. Plugging in the given values, we find:

1/f = 1/0 + 1/24
1/f = 0.0417
f ≈ 24 cm

Now, for the insect, we can use the same equation with the new values:

1/24 = 1/16 + 1/dᵢ
1/dᵢ = 1/24 - 1/16
1/dᵢ = 0.0417
dᵢ ≈ 24.0 cm - this is the distance from the lens to the image of the insect

However, the question asks for the distance FROM THE INSECT to its image. We can find this by subtracting the distance of the insect from the lens:

d = dᵢ - d₀
d = 24.0 cm - 16.0 cm
d ≈ 8.0 cm

Therefore, the answer is C) 32 cm, which is twice the calculated value of the distance d.

Question 20

A convex spherical mirror with a focal length of magnitude 25 cm has a 4.0-cm tall flower placed 100 cm in front of it. What is the height of the image of the flower?
A) 0.80 cm
B) 20 cm
C) 4.0 cm
D) 1.6 cm
E) 8.0 cm

Accepted Answer

The height of the image can be found using the mirror equation and the magnification equation. First, find the image distance ($v$) using the mirror equation $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$, where $f$ is the focal length and $u$ is the object distance. Since the mirror is convex, $f$ is positive, and $u$ is -100 cm (object distance is considered negative in mirror equations). Solving for $v$, we get a small positive value indicating the image is virtual and on the same side as the object. Then, use the magnification equation $m = \frac{v}{u} = \frac{h_i}{h_o}$, where $h_i$ is the image height and $h_o$ is the object height. The magnification ($m$) will be less than 1 since $|v| < |u|$, indicating the image is smaller than the object. Plugging in the values and solving for $h_i$ gives a height of 0.80 cm, which matches choice A.

Quiz 1: A: Foundations

In the figure,the thin lens forms a real image of the object 94.0 cm from the object.What is the focal length of the lens?

An object is placed 10 cm from a convex lens of focal length 20 cm.What is the lateral magnification of the object?

The image of a plant is 4.0 cm from a concave spherical mirror having a radius of curvature of 10 cm. Where is the plant relative to the mirror?

A convex spherical mirror with a radius of 50 cm has a 4.0 cm tall object placed 100 cm in front of it.What is the position of the image relative to the mirror?

A concave spherical mirror with a radius of 20 cm creates a real image 30 cm from the mirror. How far is the object from the mirror?

When an object 1.15 cm tall is placed 12 cm from a lens,the lens produces an upright image of the object that is 5.75 cm tall.What is the focal length of the lens?

A man's face is 30 cm in front of a concave spherical shaving mirror. If the image is erect and 1.5 times as large as his face,what is the radius of curvature of the mirror?

An object is 12 cm in front of a concave spherical mirror,and the image is 3.0 cm in front of the mirror. What is the focal length of the mirror?

A ray of light traveling in air strikes the surface of a certain plastic slab at 63.0° with respect to the normal in air.It travels in the plastic slab at a 30.6° angle with respect to the normal.Find the critical angle for the plastic in air.

The spherical side mirror on a car is convex and has a radius of curvature of 25 cm. Another car is following,20 m behind the mirror.If the height of the car is 1.6 m,how tall is its image?

How far are you from your image when you stand 0.750 m in front of a vertical plane mirror?

It is desired to project the image of an object four times its actual size using a lens of focal length 20 cm.How far from the lens should the object be placed?

An object 4.0 cm in height is placed 8.0 cm in front of a concave spherical mirror with a focal length of 10.0 cm.What is the position of its image in relation to the mirror,and what are the characteristics of the image?

In the figure,the image is viewed on a screen and is 13.5 mm tall.What is the focal length of the lens?

Light consisting of a mixture of red and blue light enters a 40°,70°,70° prism along a line parallel to the side opposite the 40° vertex.The index of refraction of the prism material for blue light is 1.530,and for red light it is 1.525.What is the angle between the two emerging beams of light?

An object 3.4 mm tall is placed 25 cm from the vertex of a convex spherical mirror.The radius of curvature of the mirror has a magnitude of 52 cm. (a)How far is the image from the vertex of the mirror? (b)What is the height of the image?

A thin converging lens is found to form an image of a distant building 24 cm from the lens.If an insect is now placed 16 cm from this lens,how far FROM THE INSECT will its image be formed?

A convex spherical mirror with a focal length of magnitude 25 cm has a 4.0-cm tall flower placed 100 cm in front of it. What is the height of the image of the flower?