Question 1

A high f-number:

Accepted Answer

A high f-number, or small aperture, results in a larger depth of focus, meaning more of the image from foreground to background will be in focus. This is useful for landscape or architectural photography where a sharp, detailed image throughout the image is desired. Higher f-numbers also result in slower shutter speeds, which can be useful in low light situations or when motion blur is desired. However, higher f-numbers can also lead to diffraction and softer images, so it is important to balance the depth of focus with the desired sharpness of the image. Spherical aberration is more commonly associated with lower f-numbers or larger apertures.

Question 2

An unclad optical fiber has index of refraction 1.40. What is the acceptance angle for this fiber if it is submerged in a fluid with index of refraction 1.30? The acceptance angle is the greatest angle at which the light enters the end of the fiber and still causes total internal reflection.

Accepted Answer

The answer of An unclad optical fiber has index of...

Question 3

A camera has a lens that has an adjustable film distance used for focusing purposes. When the object being photographed is at a very large distance, the distance from the lens to the film is 50.0 mm. What would be the lens-to-film distance for an in-focus object 2.00 m from the camera?

Accepted Answer

The lens formula is:
1/f = 1/do + 1/di
Where f is the focal length, do is the distance of the object from the lens, and di is the distance of the image from the lens. 
For the given situation, let us assume the focal length is f. 
When the object is at a very large distance, di = 50mm. This gives us: 
1/f = 1/do + 1/50
When the object is 2m away, do = 2m = 2000mm. Rearranging the above equation, we get: 
1/di = 1/f - 1/do
1/di = 1/f - 1/2000
When the object is in focus, di = do + x, where x is the lens-to-film distance. 
Substituting this in the above equation, we get: 
1/(do + x) = 1/f - 1/2000
Multiplying both sides by (do + x)2000f, we get: 
2000f(do + x) = f(do + x) - 2000(do)(do + x)
Simplifying this, we get: 
x = (2000f - dof - 50do)/(50 + do)
Substituting the values, we get: 
x = (2000f - 50f - 50*2000)/(50 + 2000) = 51.3mm
Therefore, the lens-to-film distance for an in-focus object 2.00m from the camera is 51.3mm.

Question 4

A pinhole camera is used to photograph a 23-cm tall object at a distance of 3.0 m, the image being projected on film at a distance of 46 cm from the pinhole. What are the height and orientation of the image?

Accepted Answer

The answer of A pinhole camera is used to photograph...

Question 5

The pupil of the eye increases in size under which condition?

Accepted Answer

The pupil dilates in dim light to allow more light to enter the eye, improving visibility.

Question 6

An unclad optical fiber has index of refraction 1.70. What is the acceptance angle for this fiber? The acceptance angle is the greatest angle at which the light enters the end of the fiber and still causes total internal reflection.

Accepted Answer

The answer of An unclad optical fiber has index of...

Question 7

In the NSOM technique, the letter S stands for:

Accepted Answer

NSOM stands for Near-field Scanning Optical Microscopy, where "Scanning" refers to the method of scanning a sample with a probe to obtain high-resolution images.

Question 8

An optical fiber is made with core material of index of refraction 1.6. The fiber is clad with a material of index 1.2. What is the speed of light traveling along the central axis of this fiber?

Accepted Answer

The speed of light in a medium is given by v = c/n, where c is the speed of light in vacuum (approximately 3.0 x 10^8 m/s) and n is the refractive index. For the core material with n = 1.6, the speed is v = 3.0 x 10^8 m/s / 1.6 = 1.875 x 10^8 m/s, which is approximately 1.9 x 10^8 m/s.

Question 9

A camera has a lens of focal length 50 mm and an f-number of 4.0. What is the diameter of the lens?

Accepted Answer

The answer of A camera has a lens of focal...

Question 10

A 1.5-m tall woman stands 5.0 m in front of a camera with a 8.0 cm focal length lens. What is the size of the image formed on film?

Accepted Answer

The answer of A 1.5-m tall woman stands 5.0 m...

Question 11

An optical fiber is made with core material of index of refraction 1.30. The fiber is clad with a material of index 1.50. What is the critical angle for this fiber?

Accepted Answer

The critical angle occurs when light travels from a medium with a higher index of refraction to a medium with a lower index. Here, the core has a lower index than the cladding, so total internal reflection cannot occur, and thus there is no critical angle.

Question 12

A CD is made with a plastic coating with index of refraction n and is used with a laser that produces a wavelength $\lambda$ in the plastic. Which of the following pit depths would produce the desired destructive interference in this situation?

Accepted Answer

The answer of A CD is made with a plastic...

Question 13

An optical fiber with index of refraction 1.60 is clad in a material of index 1.30. What is the ratio of the maximum time to the minimum time for light signals to travel down a given length of this fiber?

Accepted Answer

The ratio of maximum to minimum time is determined by the ratio of the refractive indices, which is 1.60/1.30 = 1.23.

Question 14

A camera is used to photograph the full moon, which has an angular diameter of 0.51°. The image on the film is 2.5 mm in diameter. What is the focal length of the lens?

Accepted Answer

We can use the formula: angular diameter = (diameter of object / distance from object to lens) x (180° / π x focal length)
Rearranging the formula: focal length = (diameter of object / distance from object to lens) x (180° / π x angular diameter)
Plugging in the given values: focal length = (2.5 mm / tan(0.51° / 2)) x (180° / π x 0.51°) = 280 mm (approx)

Question 15

A CD player uses a laser with wavelength 900 nm. If the CD has a plastic coating with index 1.50, what is the wavelength in the plastic?

Accepted Answer

The wavelength of light in a medium is given by the wavelength in vacuum divided by the refractive index of the medium. Thus, 900 nm / 1.50 = 600 nm.

Quiz 26: Applications of Optics