Question 1

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 2

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 3

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 of...

Question 4

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 5

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 final image is formed 100 cm after the second lens. This can be found using the thin lens equation:1/f = 1/do + 1/diwhere f is the focal length, do is the object distance, and di is the image distance.For the first lens, we have:1/25 = 1/100 + 1/di1Solving for di1, we get:di1 = -20 cmThis means that the first lens forms a virtual image 20 cm in front of the lens.For the second lens, we have:1/33.33 = 1/30 + 1/di2Solving for di2, we get:di2 = 100 cmThis means that the second lens forms a real image 100 cm after the lens.

Question 6

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 of...

Question 7

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 8

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? (n_water = 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 can be found using Snell's law, which relates the apparent depth to the real depth through the refractive index of water. The formula is Real Depth = Apparent Depth * Refractive Index. Therefore, the real depth = 2.00 m * 1.33 = 2.66 m.

Question 9

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 10

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 of the lens, do is the distance of the object from the lens, and di is the distance of the image from the lens.

Given: do = 16 cm, di = ?, f = ?

Since the object is placed closer to the lens than the image, we know that the lens is acting as a magnifying glass, and therefore it is a converging lens. This means that the focal length is positive.

To find the focal length, we can use the fact that the lens forms an image of a distant building 24 cm away. When the object is at infinity, the image is formed at the focal point of the lens.

1/f = 1/do + 1/di
1/f = 0 + 1/24
f = 24 cm

Now we can use the thin lens formula again to find the distance of the image from the lens when the object is at 16 cm.

1/24 = 1/16 + 1/di
di = 32 cm

Therefore, the image of the insect will be formed 32 cm from the lens, so the answer is (C) 32 cm.

Question 11

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

The answer of In the figure,the image is viewed on...

Question 12

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 13

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 14

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

The fish appears to be 2.5 cm inside the bowl. This can be found using the formula for the apparent depth of an object in water:
apparent depth = real depth / refractive index of water
In this case, the real depth is 3.0 cm (since the fish is 3.0 cm from the wall of the bowl) and the refractive index of water is 1.33. Therefore:
apparent depth = 3.0 cm / 1.33 = 2.25 cm
However, the observer is outside the bowl, so we need to add the radius of the bowl (8.0 cm) to get the final answer:
apparent position of fish = 2.25 cm + 8.0 cm = 10.25 cm
Since the fish is inside the bowl, we subtract the radius of the bowl again to get the final answer:
apparent position of fish = 10.25 cm - 8.0 cm = 2.25 cm inside the bowl
Rounded to one decimal place, this is 2.5 cm.

Question 15

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 16

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

First, we can use the thin lens equation to find the image distance for the diverging lens:
1/f = 1/di - 1/do
1/(-25) = 1/di - 1/100
di = -50 cm
(Note that the negative sign indicates that the image is virtual and upright.)

Next, we can use the thin lens equation again to find the final image distance for the whole system:
1/f = 1/di - 1/do
1/33.33 = 1/-50 - 1/30
di = -20 cm
(Note that the negative sign again indicates a virtual and upright image.)

Finally, we can use the magnification equation to find the lateral magnification:
m = -di / do
m = -(-20) / 100
m = 0.2

But we need to be careful with the sign of the magnification. Since the object is upright, a positive magnification would mean that the image is also upright (but smaller), while a negative magnification would mean that the image is inverted. In this case, the magnification is less than 1 in magnitude, so the image is smaller than the object.
Therefore, the correct answer is (C) -0.40, which indicates a virtual, upright, and smaller image.

Question 17

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 18

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 19

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 20

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 left...

Quiz 34: Geometric Optics

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?

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?

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

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?

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?

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 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)

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 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?

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 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 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 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 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?

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 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 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 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? (n_water = 1.33)