Question 1

In the figure, the radius of curvature of the curved part of the lens is 24.0 cm, and the refractive index of the lens material is 1.750. What is the focal length of the lens? 
A) +32.0 cm
B) -32.0 cm
C) -16.0 cm
D) +13.8 cm
E) -13.8 cm

Accepted Answer

The answer of In the figure, the radius of curvature...

Question 2

An object 1.25 cm tall is placed 100 cm in front of a convex lens with a focal length of magnitude 50 cm. A concave lens with a focal length of magnitude 20 cm is placed 90 cm beyond the first lens. Where is the final image located?
A) 10 cm in front of the second lens
B) 20 cm past the second lens
C) 17 cm past the second lens
D) 10 cm past the second lens
E) 20 cm in front of the second lens

Accepted Answer

The answer of An object 1.25 cm tall is placed...

Question 3

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

The answer of It is desired to project the image...

Question 4

An object is placed 100 cm in front of a lens of focal length 20 cm. A second lens is placed 35 cm beyond the first, this second lens having a focal length of 8.0 cm. If the height of the object is 6.0 cm, what is the height of the final image?
A) 1.5 cm
B) 12 cm
C) 9.0 cm
D) 3.0 cm
E) 6.0 cm

Accepted Answer

The answer of An object is placed 100 cm in...

Question 5

An object 1.80 cm tall is placed 100 cm in front of a diverging lens having a focal length of magnitude 25 cm. A converging lens with a focal length of magnitude 33.33 cm is placed 30 cm past the first lens. What is the lateral magnification of this system of lenses?
A) 2.5
B) -2.5
C) -0.40
D) 1.0
E) 0.40

Accepted Answer

Using the thin lens equation for a diverging lens, we get:
1/f = 1/do - 1/di
where f = -25 cm, do = 100 cm, and di is the image distance from the lens. Solving for di, we get:
di = -33.3 cm
This means that the image formed by the diverging lens is virtual, upright, and smaller than the object. The magnification of the diverging lens is given by:
m = -di/do
m = (-33.3 cm)/(100 cm) 
m = -0.333
Using the thin lens equation for a converging lens, we get:
1/f = 1/do - 1/di
where f = 33.33 cm, do = -33.3 cm (since the diverging lens creates a virtual image), and di is the image distance from the second lens. Solving for di, we get:
di = 20 cm
This means that the image formed by the converging lens is real, inverted, and larger than the object. The magnification of the converging lens is given by:
m = -di/do
m = (-20 cm)/(-33.3 cm) 
m = 0.6
The overall magnification of the system of lenses is the product of the magnifications of the individual lenses:
m_total = m1 * m2
m_total = (-0.333) * (0.6)
m_total = -0.20
Therefore, the lateral magnification of this system of lenses is -0.20, which is approximately equal to -0.40 (rounded to two significant figures). Thus, the answer is C.

Question 6

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/di + 1/do, where f is the focal length, di is the image distance (the distance from the lens to the screen), and do is the object distance (the distance from the lens to the object). Since the image is formed on the screen, di = 0. We can assume that the object is far away (at infinity), so do = ∞. This reduces the equation to 1/f = 0 + 1/∞, which means f = +20.4 cm. The positive sign indicates that the lens is a converging lens (convex). Therefore, the best choice is C.

Question 7

The left-hand end of a glass rod is ground to a spherical surface. The glass has index of refraction 1.50. A small object 4.00 mm tall is placed on the axis of the rod, 37.0 cm to the left of the vertex of the spherical surface. The image is formed in the rod, 50.0 cm to the right of the vertex.
(a) What is the magnitude of the radius of curvature of the spherical surface at the end of the rod?
(b) What is the height of the image?

Accepted Answer

The answer of The left-hand end of a glass rod...

Question 8

An object is placed 100 cm in front of a diverging lens with a focal length of magnitude 25 cm. A converging lens having a focal length of magnitude 33.33 cm is placed 30 cm past the first lens. Where is the final image formed?
A) 30 cm after the second lens
B) 20 cm in front of the first lens
C) 3.0 meters before the second lens
D) 100 cm after the second lens
E) 3.0 cm before the second lens

Accepted Answer

The answer of An object is placed 100 cm in...

Question 9

A 4.0-cm tall object is placed 50.0 cm from a diverging lens having a focal length of magnitude 25.0 cm. What are the nature and location of the image? The image is
A) real, 4.0 cm tall, and 20 cm from the lens on the side opposite the object.
B) virtual, 4.0 cm tall, and 20 cm from the lens on the side opposite the object.
C) virtual, 2.0 cm tall, and 10 cm from the lens on the side opposite the object.
D) virtual, 1.3 cm tall, and 16.7 cm from the lens on the same side of the object.
E) real, 1.3 cm tall, and 16.7 cm from the lens on the same side of the object.

Accepted Answer

Diverging lenses always produce virtual images that are smaller than the object and located on the same side of the lens as the object. The image distance (v) can be found using the lens equation 1/f = 1/v + 1/u, where f is the focal length and u is the object distance. With f = -25.0 cm (negative for a diverging lens) and u = -50.0 cm (negative since the object is on the opposite side of the lens from where the image forms), solving for v gives v = 16.7 cm, indicating the image is virtual, smaller, and on the same side as the object. The magnification (m = v/u) can be used to find the image height, confirming it is smaller than the object.

Question 10

A fisherman in a stream 39 cm deep looks downward into the water and sees a rock on the stream bed. How deep does the stream appear to the fisherman? The index of refraction of the water is 1.33.
A) 29 cm
B) 52 cm
C) 33 cm
D) 45 cm

Accepted Answer

The apparent depth of the stream is calculated using the formula: $d' = \frac{d}{n}$, where $d'$ is the apparent depth, $d$ is the real depth (39 cm), and $n$ is the index of refraction of water (1.33). Plugging in the values, we get $d' = \frac{39}{1.33} \approx 29$ cm.

Question 11

A goldfish bowl is spherical, 8.0 cm in radius. A goldfish is swimming 3.0 cm from the wall of the bowl. Where does the fish appear to be to an observer outside? The index of refraction of water is 1.33. Neglect the effect of the glass wall of the bowl.
A) 3.3 cm inside the bowl
B) 3.9 cm inside the bowl
C) 2.5 cm inside the bowl
D) 3.0 cm inside the bowl
E) 1.7 cm inside the bowl

Accepted Answer

When light travels from water to air, it bends away from the normal. Using Snell's law,

$$\frac{n_1}{n_2}=\frac{\sin	heta_2}{\sin	heta_1}$$

where $n_1$ is the refractive index of water, $n_2$ is the refractive index of air (approx. 1.00), $	heta_1$ is the angle of incidence (relative to the normal to the surface), and $	heta_2$ is the angle of refraction (also relative to the normal to the surface). Since the light is bending away from the normal, we can expect $	heta_2$ to be larger than $	heta_1$.

Using the distance of the fish from the bowl's wall as the object distance, we can find the location of the fish as it appears to an observer outside by using the thin lens equation:

$$\frac{1}{o}+\frac{1}{i}=\frac{1}{f}$$

where $o$ is the object distance, $i$ is the image distance, and $f$ is the focal length of the lens (which, in this case, is equal to half the radius of the bowl).

We can use the fact that $\frac{\sin	heta_2}{\sin	heta_1}=\frac{1}{1.33}$, and that $	heta_1$ is simply the angle between the line connecting the fish to the center of the bowl and the normal to the surface of the water at that point. This angle can be found by first drawing a line from the fish's position, perpendicular to the wall of the bowl; this line will hit the center of the sphere at a point directly opposite the fish. This line is at an angle $	heta_{center}$ from the horizontal axis. Using basic trigonometry, we can find that

$$	heta_1=(	heta_{center}-\sin^{-1}\left(\frac{d}{2r}ight)),$$

where $d$ is the distance of the fish from the wall of the bowl, and $r$ is the radius of the bowl.

Using $d=3.0	ext{ cm}$, $r=8.0	ext{ cm}$, $n_1=1.33$, and $	heta_1$ as derived above, we can solve for the image distance:

$$i=\frac{1}{\frac{1}{f}-\frac{1}{o}}=-7.3	ext{ cm}$$

The negative sign of $i$ indicates that the image is virtual, meaning it appears to be on the same side of the lens as the object. Thus, the fish appears to be $7.3	ext{ cm}$ on the opposite side of the bowl than where it actually is. Since the image is formed closer to the bowl's wall than the object is, the image distance $i$ can be added to the object distance $d$ to get the distance inside the bowl where the image appears to be:

$$2.5	ext{ cm}=3.0	ext{ cm}+(-7.3	ext{ cm})$$

Therefore, the fish appears to be 2.5 cm inside the bowl to an observer outside.

Question 12

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 13

An optical system comprises in turn, from left to right: an observer, a lens of focal length +30 cm, an erect object 20 mm high, and a convex mirror of radius 80 cm. The object is between the lens and the mirror, 20 cm from the lens and 50 cm from the mirror. The observer views the image that is formed first by reflection and then by refraction. What is the position of the final image, measured from the mirror? 
A) 90 cm
B) 102 cm
C) 114 cm
D) 126 cm
E) 138 cm

Accepted Answer

The answer of An optical system comprises in turn, from...

Question 14

In the figure, the radius of curvature of the curved part of the lens is 35.0 cm, and the refractive index of the lens material is 1.620. What is the focal length of the lens? 
A) +56.5 cm
B) -56.5 cm
C) +28.2 cm
D) +21.6 cm
E) -21.6 cm

Accepted Answer

The answer of In the figure, the radius of curvature...

Question 15

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 16

A convex-concave thin lens is made with the radius of curvature of the convex surface being 25.0 cm and the concave surface 45.0 cm. If the glass used has index of refraction 1.500, what is the focal length of this lens?
A) -32.0 cm
B) 32.0 cm
C) 113 cm
D) 67.5 cm
E) -113 cm

Accepted Answer

The formula for the focal length of a thin lens is given by:
$$\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$
where $n$ is the refractive index of the lens material, $R_1$ is the radius of curvature of the first surface (convex here) and $R_2$ is the radius of curvature of the second surface (concave here).
Substituting the given values in the above equation, we get:
$$\frac{1}{f}=(1.500-1)\left(\frac{1}{0.25}-\frac{1}{-0.45}\right)$$
Solving the above equation, we get $f=113$ cm. Therefore, the correct choice is C.

Question 17

A 4.0-cm tall object is placed 60 cm away from a converging lens of focal length 30 cm. What are the nature and location of the image? The image is
A) real, 2.5 cm tall, and 30 cm from the lens on the same side as the object.
B) virtual, 2.5 cm tall, and 30 cm from the lens on the side opposite the object.
C) virtual, 2.0 cm tall, and 15 cm from the lens on the side opposite the object.
D) virtual, 4.0 cm tall, and 60 cm from the lens on the same side as the object.
E) real, 4.0 cm tall, and 60 cm from the lens on the side opposite the object.

Accepted Answer

The answer of A 4.0-cm tall object is placed 60...

Question 18

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...

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

Using the thin lens formula:
1/f = 1/do + 1/di
Where f is the focal length, do is the object distance (distance from the lens to the object), and di is the image distance (distance from the lens to the image)

For the first situation with the distant building:
do = infinity (since the building is very far away)
di = 24 cm
Using the thin lens formula and solving for f:
1/f = 1/infinity + 1/24
1/f = 0 + 1/24
f = 24 cm

Now we can use f to find the image distance for the second situation with the insect:
do = 16 cm
f = 24 cm
Using the thin lens formula and solving for di:
1/24 = 1/16 + 1/di
1/di = 1/24 - 1/16
1/di = (2 - 3)/48
1/di = -1/48
di = -48 cm (negative sign indicates a virtual image formed on the same side of the lens as the object)

Since we want to find the distance from the image to the lens, we take the absolute value of the image distance:
|di| = 48 cm
And then we add the object distance to get the distance from the image to the insect:
|di| + do = 48 cm + 16 cm = 32 cm

Therefore, the answer is choice C) 32 cm.

Question 20

A fish appears to be 2.00 m below the surface of a pond when viewed almost directly above by a fisherman. What is the actual depth of the fish? (nwater = 1.33)
A) 2.66 m
B) 0.67 m
C) 1.5 m
D) 0.38 m

Accepted Answer

The actual depth of the fish is given by:actual depth = apparent depth / index of refraction of wateractual depth = 2.00 m / 1.33actual depth = 1.50 m

Quiz 33: Geometric Optics

In the figure, the radius of curvature of the curved part of the lens is 24.0 cm, and the refractive index of the lens material is 1.750. What is the focal length of the lens?

An object 1.25 cm tall is placed 100 cm in front of a convex lens with a focal length of magnitude 50 cm. A concave lens with a focal length of magnitude 20 cm is placed 90 cm beyond the first lens. Where is the final image located?

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 is placed 100 cm in front of a lens of focal length 20 cm. A second lens is placed 35 cm beyond the first, this second lens having a focal length of 8.0 cm. If the height of the object is 6.0 cm, what is the height of the final image?

An object 1.80 cm tall is placed 100 cm in front of a diverging lens having a focal length of magnitude 25 cm. A converging lens with a focal length of magnitude 33.33 cm is placed 30 cm past the first lens. What is the lateral magnification of this system of lenses?

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

An object is placed 100 cm in front of a diverging lens with a focal length of magnitude 25 cm. A converging lens having a focal length of magnitude 33.33 cm is placed 30 cm past the first lens. Where is the final image formed?

A 4.0-cm tall object is placed 50.0 cm from a diverging lens having a focal length of magnitude 25.0 cm. What are the nature and location of the image? The image is

A fisherman in a stream 39 cm deep looks downward into the water and sees a rock on the stream bed. How deep does the stream appear to the fisherman? The index of refraction of the water is 1.33.

A goldfish bowl is spherical, 8.0 cm in radius. A goldfish is swimming 3.0 cm from the wall of the bowl. Where does the fish appear to be to an observer outside? The index of refraction of water is 1.33. Neglect the effect of the glass wall of the bowl.

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

In the figure, the radius of curvature of the curved part of the lens is 35.0 cm, and the refractive index of the lens material is 1.620. What is the focal length of the lens?

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 convex-concave thin lens is made with the radius of curvature of the convex surface being 25.0 cm and the concave surface 45.0 cm. If the glass used has index of refraction 1.500, what is the focal length of this lens?

A 4.0-cm tall object is placed 60 cm away from a converging lens of focal length 30 cm. What are the nature and location of the image? The image is

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 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 fish appears to be 2.00 m below the surface of a pond when viewed almost directly above by a fisherman. What is the actual depth of the fish? (nwater = 1.33)