Question 1

Two converging lenses, the first with focal length $10.0 \mathrm{~cm}$ and the second with focal length $20.0 \mathrm{~cm}$ , are separated by $40.0 \mathrm{~cm}$ . An object, $3.00 \mathrm{~cm}$ in height, is placed $30.0 \mathrm{~cm}$ in front of the first lens. Where is the final image formed?

A)

45.0 \mathrm{~cm}

after the second lens
B)

100 \mathrm{~cm}

after the second lens
C)

5.00 \mathrm{~cm}

after the second lens
D)

11.1 \mathrm{~cm}

after the second lens
E)

5.00 \mathrm{~cm}

before the second lens

100 \mathrm{~cm}

after the second lens

Accepted Answer

First, calculate the image position from the first lens using the lens formula. Then, treat this image as the object for the second lens and calculate the final image position.

Question 2

Two converging lenses, the first with focal length $10.0 \mathrm{~cm}$ and the second with focal length $20.0 \mathrm{~cm}$ , are separated by $40.0 \mathrm{~cm}$ . An object, $3.00 \mathrm{~cm}$ in height, is placed $30.0 \mathrm{~cm}$ in front of the first lens. Where is the final image formed?

A)

45.0 \mathrm{~cm}

after the second lens
B)

100 \mathrm{~cm}

after the second lens
C)

5.00 \mathrm{~cm}

after the second lens
D)

11.1 \mathrm{~cm}

after the second lens
E)

5.00 \mathrm{~cm}

before the second lens

100 \mathrm{~cm}

after the second lens

Accepted Answer

To find the image location formed by the first lens, we use the lens 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. Plugging in the values for the first lens ($f = 10.0 \, \text{cm}$, $d_o = 30.0 \, \text{cm}$), we get $\frac{1}{10} = \frac{1}{30} + \frac{1}{d_i}$, solving for $d_i$ gives $d_i = 15.0 \, \text{cm}$. This means the image is formed $15.0 \, \text{cm}$ after the first lens.

Question 3

Two converging lenses, the first with focal length $10.0 \mathrm{~cm}$ and the second with focal length $20.0 \mathrm{~cm}$ , are separated by $40.0 \mathrm{~cm}$ . An object, $3.00 \mathrm{~cm}$ in height, is placed $30.0 \mathrm{~cm}$ in front of the first lens. What is the height of the final image?

A)

1.50 \mathrm{~cm}

B)

12.0 \mathrm{~cm}

C)

6.00 \mathrm{~cm}

D)

3.00 \mathrm{~cm}

E)

24.0 \mathrm{~cm}

6.00 \mathrm{~cm}

Accepted Answer

First, find the image distance from the first lens using the lens formula: $ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} $. For the first lens, $ f = 10 \, \text{cm} $ and $ d_o = 30 \, \text{cm} $, so $ d_i = 15 \, \text{cm} $. The image from the first lens acts as the object for the second lens, located $ 40 - 15 = 25 \, \text{cm} $ from the second lens. Using the lens formula again for the second lens with $ f = 20 \, \text{cm} $ and $ d_o = 25 \, \text{cm} $, we find $ d_i = 100 \, \text{cm} $. The magnification from the first lens is $ -\frac{d_i}{d_o} = -\frac{15}{30} = -0.5 $, and from the second lens is $ -\frac{d_i}{d_o} = -\frac{100}{25} = -4 $. The total magnification is $ (-0.5) \times (-4) = 2 $. Therefore, the final image height is $ 3 \, \text{cm} \times 2 = 6 \, \text{cm} $.

Question 4

Two converging lenses, the first with focal length $10.0 \mathrm{~cm}$ and the second with focal length $20.0 \mathrm{~cm}$ , are separated by $40.0 \mathrm{~cm}$ . An object, $3.00 \mathrm{~cm}$ in height, is placed $30.0 \mathrm{~cm}$ in front of the first lens. What is the height of the final image?

A)

1.50 \mathrm{~cm}

B)

12.0 \mathrm{~cm}

C)

6.00 \mathrm{~cm}

D)

3.00 \mathrm{~cm}

E)

24.0 \mathrm{~cm}

6.00 \mathrm{~cm}

Accepted Answer

The answer of  Two converging lenses, the first with...

Question 5

Two lenses, the first with focal length $10.0 \mathrm{~cm}$ and the second with focal length $-12.5 \mathrm{~cm}$ , are separated by $20.0 \mathrm{~cm}$ . An object of height $2.40 \mathrm{~cm}$ is placed $15.0 \mathrm{~cm}$ in front of the first lens. Where is the final image?

A)

12.5 \mathrm{~cm}

after the second lens
B)

20.0 \mathrm{~cm}

after the second lens
C)

100 \mathrm{~cm}

after the second lens
D)

50.0 \mathrm{~cm}

after the second lens
E)

5.60 \mathrm{~cm}

after the second lens

Accepted Answer

The answer of  Two lenses, the first with focal...

Question 6

Two lenses, the first with focal length $10.0 \mathrm{~cm}$ and the second with focal length $-12.5 \mathrm{~cm}$ , are separated by $20.0 \mathrm{~cm}$ . An object of height $2.40 \mathrm{~cm}$ is placed $15.0 \mathrm{~cm}$ in front of the first lens. What is the size of the final image?

A)

4.80 \mathrm{~cm}

B)

12.4 \mathrm{~cm}

C)

2.40 \mathrm{~cm}

D)

24.0 \mathrm{~cm}

E)

0.96 \mathrm{~cm}

Accepted Answer

The answer of  Two lenses, the first with focal...

Question 7

Two lenses, the first with focal length $10.0 \mathrm{~cm}$ and the second with focal length $-12.5 \mathrm{~cm}$ , are separated by $20.0 \mathrm{~cm}$ . An object of height $2.40 \mathrm{~cm}$ is placed $15.0 \mathrm{~cm}$ in front of the first lens. Which of the following characteristics does the final image have?

A) virtual and upright
B) real and upright
C) virtual and diminished in size
D) virtual and inverted
E) real and inverted

Accepted Answer

The first lens, being positive, acts as a converging lens and forms a real and inverted image. This image then acts as the object for the second lens, which is negative and thus a diverging lens. The combination of these effects results in a final image that is real and inverted.

Question 8

A $50.0 \mathrm{~mm}$ lens is used to take a picture of an object $1.30 \mathrm{~m}$ tall located $4.00 \mathrm{~m}$ away. What is the height of the image on the film?&#10;A) $16.5 \mathrm{~mm}$&#10;B) $21.0 \mathrm{~mm}$&#10;C) $10.5 \mathrm{~mm}$&#10;D) $12.0 \mathrm{~mm}$&#10;E) $9.20 \mathrm{~mm}$

Accepted Answer

The height of the image can be calculated using the lens formula and magnification. The magnification (m) is given by the ratio of the image height (h') to the object height (h), and also equals the ratio of the image distance (v) to the object distance (u). Given that the object distance (u) is $4.00 \mathrm{~m}$ and the focal length (f) of the lens is $50.0 \mathrm{~mm} = 0.050 \mathrm{~m}$, we can first find the image distance (v) using the lens equation $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$. Solving for v gives us the image distance. Then, using the magnification equation $m = \frac{h'}{h} = \frac{v}{u}$, we can solve for the image height (h'). The correct calculation leads to an image height of approximately $16.5 \mathrm{~mm}$, making option A correct.

Question 9

A nearsighted person wears contact lenses of refractive power $-6.0 \mathrm{D}$. What is his far point?&#10;A) $-17 \mathrm{~cm}$&#10;B) $-6.0 \mathrm{~cm}$&#10;C) $6.0 \mathrm{~cm}$&#10;D) $-4.0 \mathrm{~cm}$&#10;E) $17 \mathrm{~cm}$

Accepted Answer

The far point of a nearsighted person is the maximum distance at which they can see objects clearly without correction. The refractive power of the lenses needed for correction is given by $P = -\frac{1}{f}$, where $P$ is the power in diopters and $f$ is the focal length in meters. Given $P = -6.0 \mathrm{D}$, we find $f = -\frac{1}{6} \mathrm{m} = -16.67 \mathrm{cm}$, which is approximately $17 \mathrm{cm}$ but as a positive distance, since the far point conceptually refers to a positive distance in front of the eyes.

Question 10

A concave mirror has a radius of $2.0 \mathrm{~m}$. What is the refractive power of a lens of the same focal length as the mirror?&#10;A) $0.75 \mathrm{D}$&#10;B) $-0.50 \mathrm{D}$&#10;C) $2.0 \mathrm{D}$&#10;D) $1.0 \mathrm{D}$&#10;E) $0.50 \mathrm{D}$

Accepted Answer

The answer of A concave mirror has a radius of...

Question 11

The refractive power of a lens of focal length $40 \mathrm{~cm}$ is&#10;A) $-2.5 \mathrm{D}$.&#10;B) $2.5 \mathrm{D}$.&#10;C) $0.40 \mathrm{D}$.&#10;D) $0.16 \mathrm{D}$.&#10;E) $4.0 \mathrm{D}$.

Accepted Answer

The answer of  The refractive power of a lens...

Question 12

A lens has a refractive power of $-1.50 \mathrm{D}$. What is its focal length?&#10;A) $8.50 \mathrm{~cm}$&#10;B) $66.7 \mathrm{~cm}$&#10;C) $-66.7 \mathrm{~cm}$&#10;D) $150 \mathrm{~cm}$&#10;E) $-150 \mathrm{~cm}$

Accepted Answer

The answer of A lens has a refractive power of...

Question 13

For maximum angular magnification in a compound microscope, the objective focal length $\left(f_{o}\right)$ and the eyepiece focal length $\left(f_{e}\right)$ should be chosen so that&#10;A) $f_{o}$ and $f_{e}$ are both the smallest available.&#10;B) $f_{o}$ and $f_{e}$ are about the same.&#10;C) $f_{o}$ and $f_{e}$ are both the largest available.&#10;D) $f_{o}$ is the largest available; $f_{e}$ is the smallest available.&#10;E) $f_{o}$ is the smallest available; $f_{e}$ is the largest available.

Accepted Answer

The answer of For maximum angular magnification in a compound...

Question 14

An astronomical telescope is made with lenses of focal lengths 80.0 and $2.00 \mathrm{~cm}$. What is the barrel length of this telescope?&#10;A) $105 \mathrm{~cm}$&#10;B) $82.0 \mathrm{~cm}$&#10;C) $80.0 \mathrm{~cm}$&#10;D) $78.0 \mathrm{~cm}$&#10;E) $127 \mathrm{~cm}$

Accepted Answer

The answer of An astronomical telescope is made with lenses...

Deck 24: Optical Instruments