Question 1

The wavelength of green light in air is 500 nm.The wavelength of this light in glass of index of refraction 1.60 is approximately
A)313 nm
B)500 nm
C)750 nm
D)800 nm
E)427 nm

Accepted Answer

The wavelength of light in a medium is given by the wavelength in vacuum divided by the refractive index of the medium. So, 500 nm / 1.60 = 312.5 nm, which is approximately 313 nm.

Question 2

Use the following figure below for the next problem. The figure shows a cross- -A fiber optic is made by cladding a thin fiber core of refractive index n₁ = 1.45 with a material of refractive index n₂ = 1.38.The numerical aperture of a fiber optic is defined as NA = n_air sin $\theta$_m,where $\theta$_m is the maximum incident angle so that the light ray is totally internally reflected inside the fiber.The NA for the fiber is A)0.445 B)0.614 C)0.645 D)0.407 E)0.470

Accepted Answer

The numerical aperture (NA) of a fiber optic is given by NA = n_air * sin(theta_m), where theta_m is the maximum incident angle for which light is totally internally reflected inside the fiber. In this case, n_air = 1 (since the fiber is in air), n_1 = 1.45, and n_2 = 1.38. Using Snell's law, we can find that sin(theta_m) = sqrt((n_1^2 - n_2^2) / n_1^2) = 0.445. Therefore, the NA is 0.445.

Question 3

A light wave traveling in air impinges on an amber plate at an angle of incidence of 60º.If the angle of refraction is 34º,the velocity of light in amber must be approximately A)0.52 $\times$ 10⁸ m/s B)4.64 $\times$ 10⁸ m/s C)1.86 $\times$ 10³ m/s D)2.09 $\times$ 10⁸ m/s E)4.64 $\times$ 10¹⁰ m/s

Accepted Answer

Snell's law: n1 sin θ1 = n2 sin θ2n1 = 1 (air)n2 = ? (amber)θ1 = 60ºθ2 = 34ºn2 = n1 sin θ1 / sin θ2n2 = 1 * sin 60º / sin 34ºn2 ≈ 1.86v = c / nv = 3 * 10^8 m/s / 1.86v ≈ 1.61 * 10^8 m/s

Question 4

A small beam of monochromatic light in air is incident on a large plate of glass with plane-parallel sides at an angle of 30º.The index of refraction of the glass plate for this wavelength is 1.50.The angle of emergence of the light on the far side of the plate is approximately
A)30º
B)45º
C)20º
D)the angle whose sine is 1/3.
E)60º

Accepted Answer

The answer of A small beam of monochromatic light in...

Question 5

The dependence of the index of refraction on wavelength is called dispersion.A beam of light (in air)containing red and blue light is incident at an angle of 35S1U1P1$\circ$S1S1P0 on a glass plate that has an index of refraction of 1.64 for blue light and 1.61 for red light.Calculate the angle between the red and blue light inside the glass.
A)1.3S1U1P1$\circ$S1S1P0
B)0.40S1U1P1$\circ$S1S1P0
C)0.56S1U1P1$\circ$S1S1P0
D)0.74S1U1P1$\circ$S1S1P0
E)0.13S1U1P1$\circ$S1S1P0

Accepted Answer

To find the angle between the red and blue light inside the glass, we use Snell's law, $n_1 \sin(\theta_1) = n_2 \sin(\theta_2)$, where $n_1$ and $n_2$ are the indices of refraction of the first and second medium, respectively, and $\theta_1$ and $\theta_2$ are the angles of incidence and refraction, respectively. Given that the index of refraction for blue light is 1.64 and for red light is 1.61, and the angle of incidence is 35 degrees, we can calculate the angles of refraction for both colors inside the glass. The difference between these angles gives the dispersion angle between the red and blue light inside the glass. Calculating these angles and their difference yields an angle of approximately 0.40 degrees, making option B the correct answer.

Question 6

The dependence of the index of refraction on wavelength is called dispersion.A beam of light (in air)containing red and blue light is incident at an angle of 35S1U1P1$\circ$S1S1P0 on a glass plate that has an index of refraction of 1.64 for blue light and 1.61 for red light.By what percent does the speed of the red light exceed that of the blue light inside the glass,i.e.calculate (v(red)-v(blue))* 100 / v(red)?
A)1.1%
B)1.8%
C)0.53%
D)The speed is the same for both colors.
E)0.019%

Accepted Answer

The speed of light in a medium is given by $v = \frac{c}{n}$, where $c$ is the speed of light in vacuum and $n$ is the index of refraction. For red light, $v_{\text{red}} = \frac{c}{1.61}$ and for blue light, $v_{\text{blue}} = \frac{c}{1.64}$. The percent difference in speed is calculated as $\frac{v_{\text{red}} - v_{\text{blue}}}{v_{\text{red}}} \times 100$. Given that the indices of refraction for red and blue light are 1.61 and 1.64 respectively, the speed of red light is slightly higher than that of blue light due to its lower index of refraction. Calculating the exact percentage gives a result close to 1.8%.

Question 7

Two clear but non-mixing liquids each of depth 15 cm are placed together in a glass container.The liquids have refractive indices of 1.75 and 1.33.What is the apparent depth of the combination when viewed from above?
A)19.8 cm
B)23.1 cm
C)17.7 cm
D)9.9 cm
E)None of these is correct.

Accepted Answer

To find the apparent depth, we use the formula:

Apparent depth = Actual depth / Refractive index

Let the actual depths of the two liquids be h1 and h2, respectively, where h1 + h2 = 15 cm (since each liquid has a depth of 15 cm).

Then, the apparent depth of the first liquid when viewed from above is:

h1' = h1 / n1

Similarly, the apparent depth of the second liquid is:

h2' = h2 / n2

When viewed from above, the apparent depth of the combination is the sum of the two apparent depths:

h' = h1' + h2'

Substituting the values, we get:

h' = (h1 / n1) + (h2 / n2)

h' = (15 / 1.75) + (15 / 1.33)

h' = 8.57 + 11.28

h' = 19.85 cm (approx)

Therefore, the apparent depth of the combination when viewed from above is 19.8 cm(approx).

Question 8

An "X" is marked on the bottom of a glass container.A microscope is adjusted so that it is focused on the "X." A liquid is now poured into the glass to a depth of 6.00 cm.If the microscope has to be backed up by a distance of 1.50 cm to refocus on the "X," then calculate the refractive index of the liquid.
A)1.25
B)4.00
C)1.33
D)0.75
E)3.00

Accepted Answer

The refractive index of the liquid can be calculated using the formula:

n = d / (d - h)

where, 
n = refractive index of the liquid
d = depth of the liquid = 6.00 cm
h = distance the microscope was backed up = 1.50 cm

Substituting the values, we get:

n = 6.00 / (6.00 - 1.50) = 1.33

Therefore, the refractive index of the liquid is 1.33. The correct answer is choice C.

Question 9

A 60S1U1P1$\circ$S1S1P0 prism has a refractive index of 1.64 for blue light and 1.61 for red light.Calculate the difference in the angle of emergence from the opposite face of the prism for a beam of red and a beam of blue light that are both incident at an angle of 60^o. A)0.4S1U1P1$\circ$S1S1P0 B)0.0S1U1P1$\circ$S1S1P0 C)1.3S1U1P1$\circ$S1S1P0 D)3.9S1U1P1$\circ$S1S1P0 E)2.7S1U1P1$\circ$S1S1P0

Accepted Answer

The answer of A 60S1U1P1$\circ$S1S1P0 prism has a refractive index...

Question 10

Use the following figure below for the next problem. The figure shows a cross- -A fiber optic is made by cladding a thin fiber core of refractive index n₁ = 1.45 with a material of refractive index n₂ = 1.38.What is the critical angle,$\theta$_c,at the core-cladding interface? A)44º B)46º C)55º D)72º E)None of these is correct.

Accepted Answer

The answer of Use the following figure below for the...

Question 11

A 60S1U1P1$\circ$S1S1P0 prism has a refractive index of 1.52.Calculate the angle of incidence for the initial light ray that is then refracted inside the prism and just undergoes total internal reflection at the opposite face of the prism.
A)41.1S1U1P1$\circ$S1S1P0
B)18.9S1U1P1$\circ$S1S1P0
C)29.4S1U1P1$\circ$S1S1P0
D)30.6S1U1P1$\circ$S1S1P0
E)60.6S1U1P1$\circ$S1S1P0

Accepted Answer

For total internal reflection to occur at the opposite face, the angle of incidence inside the prism at that face must equal the critical angle. The critical angle can be found using Snell's law, $n_1 \sin(\theta_c) = n_2 \sin(90^\circ)$, where $n_1$ is the refractive index of the prism (1.52), $n_2$ is the refractive index of air (approximately 1), and $\theta_c$ is the critical angle. Solving for $\theta_c$ gives a value that, when applied to the geometry of the prism and using Snell's law again for the entrance face, results in an initial angle of incidence that allows the light to just undergo total internal reflection at the opposite face, which corresponds to the scenario described.

Question 12

A light wave in air strikes a glass plate 1 cm thick at an angle of incidence of 50.0º.If the index of refraction of the glass is 1.50 for this light,the actual path length of the light through the glass is approximately
A)0.51 cm
B)1.16 cm
C)1.50 cm
D)2.11 cm
E)7.66 cm

Accepted Answer

The actual path length of the light through the glass is given by the product of the thickness of the glass and the sine of the angle of refraction. The angle of refraction can be found using Snell's law:n1 * sin(theta1) = n2 * sin(theta2)where n1 is the index of refraction of the first medium (air), theta1 is the angle of incidence, n2 is the index of refraction of the second medium (glass), and theta2 is the angle of refraction.Plugging in the given values, we get:1 * sin(50.0º) = 1.50 * sin(theta2)Solving for theta2, we get:theta2 = 32.4ºTherefore, the actual path length of the light through the glass is:1 cm * sin(32.4º) = 0.51 cmTherefore, the correct answer is A.

Question 13

If the refractive index of glass is 1.5 and the refractive index of a particular liquid is 1.38,then calculate the value for the critical angle of total internal reflection at the liquid-glass interface.
A)23.1S1U1P1$\circ$S1S1P0
B)66.9S1U1P1$\circ$S1S1P0
C)82.8S1U1P1$\circ$S1S1P0
D)42.6S1U1P1$\circ$S1S1P0
E)None of these is correct.

Accepted Answer

The critical angle ($ \theta_c $) for total internal reflection at the interface between two media can be calculated using Snell's law, $ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) $, where $ n_1 $ and $ n_2 $ are the refractive indices of the two media, and $ \theta_1 $ and $ \theta_2 $ are the angles of incidence and refraction, respectively. For total internal reflection, $ \theta_2 = 90^\circ $, so $ \sin(\theta_2) = 1 $, and $ \theta_1 = \theta_c $, the critical angle. Rearranging for $ \theta_c $, we get $ \sin(\theta_c) = \frac{n_2}{n_1} $. Substituting $ n_1 = 1.38 $ (liquid) and $ n_2 = 1.5 $ (glass), $ \sin(\theta_c) = \frac{1.5}{1.38} $. Solving for $ \theta_c $ gives a value that, when calculated, is closest to 66.9 degrees, indicating that option B is correct.

Question 14

You shine light of frequency 4.00 $\times$ 10¹⁴Hz through a glass plate whose index of refraction for this light is 1.50.The wavelength of this light in the glass is approximately A)500 nm B)110 nm C)800 nm D)425 nm E)380 nm

Accepted Answer

The formula to find the wavelength of light in a material with refractive index n is:

λ = λ0/n

where λ0 is the wavelength of light in vacuum.

In this case, the frequency of light is given as 4.00 × 10^14 Hz, which corresponds to a wavelength of:

λ0 = c/f = 3.00 × 10^8 m/s / 4.00 × 10^14 Hz = 7.50 × 10^-7 m = 750 nm

where c is the speed of light.

Using the formula above, we can find the wavelength of light in the glass:

λ = λ0/n = 750 nm / 1.50 = 500 nm

Therefore, the correct answer is C.

Question 15

The minimum angle of deviation produced by a glass prism (n = 1.70 for the light used)whose apex angle is 45º is approximately
A)32º
B)36º
C)26º
D)18º
E)45º

Accepted Answer

The minimum angle of deviation ($\delta_{min}$) for a prism occurs when the light ray passes through the prism symmetrically. This condition is given by the formula $\delta_{min} = i - A$, where $i$ is the angle of incidence and $A$ is the apex angle of the prism. For minimum deviation, the angle of incidence equals the angle of emergence, and Snell's law gives $n = \frac{\sin\left(\frac{A + \delta_{min}}{2}\right)}{\sin\left(\frac{A}{2}\right)}$. Substituting the given values $n = 1.70$ and $A = 45^\circ$, and solving for $\delta_{min}$ yields a value close to 32º.

Question 16

A column of liquid is 10 cm high.Viewed from above the liquid has an apparent depth of 6.77 cm.If the refractive index of the liquid is 1.33 at the top and increases uniformly to the bottom,then what is the refractive index at the bottom of the liquid?
A)1.96
B)2.95
C)1.48
D)1.62
E)None of these is correct.

Accepted Answer

Using the formula for apparent depth, we can write:
Real depth = Apparent depth / Refractive index
Let the real depth of the liquid be h.
h = 6.77 / 1.33
h = 5.08 cm
Now, let's consider a small element of height dh at a depth h from the top. The refractive index at this depth can be written as:
n(h) = 1.33 + k.h
where k is a constant.
We know that at the top (h = 0), n = 1.33. Plugging this in the above equation, we get:
1.33 = 1.33 + k.0
k = 0
So, the refractive index equation becomes:
n(h) = 1.33
This means that the refractive index is constant throughout the liquid, and hence equal to 1.33.

Question 17

Use the following figure below for the next problem. The figure shows a cross- -A fiber optic is made by cladding a thin fiber core of refractive index n₁ = 1.45 with a material of refractive index n₂ = 1.38.What is the maximum incident angle,$\theta$,so that the light ray is totally internally reflected inside the fiber? A)24.1º B)26.4º C)28.2º D)30.3º E)32.7º

Accepted Answer

Using Snell's law,
nsinθ = n1sinθ1
where n = refractive index of cladding, n1 = refractive index of core, θ = incident angle, and θ1 = angle of refraction (which will be 90° in this case for total internal reflection).
Rearranging the equation gives:
sinθ = (n1/n)sinθ1
θ = sin^(-1)((n1/n)sinθ1)
θ1 is the critical angle, which is the maximum incident angle for total internal reflection:
sinθ1 = n/n1
Substituting the values given,
sinθ1 = 1.45/1.38
sinθ1 = 1.0507
θ1 = sin^(-1)(1.0507)
θ1 = 82.18°
Substituting θ1 into the equation for θ,
θ = sin^(-1)((1.45/1.38)sin82.18°)
θ = 26.40° (rounded to two decimal places)
Therefore, the maximum incident angle for total internal reflection is 26.40°, which corresponds to choice B.

Question 18

A 60S1U1P1$\circ$S1S1P0 prism has a refractive index of 1.71.Calculate the angle of emergence out of the prism for a ray that enters one side at an angle almost parallel to one face,i.e.at 90S1U1P1$\circ$S1S1P0 to the normal.
A)24.3S1U1P1$\circ$S1S1P0
B)35.8S1U1P1$\circ$S1S1P0
C)45.1S1U1P1$\circ$S1S1P0
D)48.7S1U1P1$\circ$S1S1P0
E)55.3S1U1P1$\circ$S1S1P0

Accepted Answer

The answer of A 60S1U1P1$\circ$S1S1P0 prism has a refractive index...

Question 19

The critical angle for light rays passing from flint glass (n = 1.70)into air is about
A)3.4º
B)36º
C)44º
D)54º
E)90º

Accepted Answer

The critical angle can be calculated using Snell's Law, where n1*sin(theta_c) = n2*sin(90 degrees), with n1 being the refractive index of the initial medium (flint glass, n = 1.70) and n2 being the refractive index of the second medium (air, n ≈ 1.00). Solving for the critical angle (theta_c) gives a value close to 36 degrees.

Question 20

A 60S1U1P1$\circ$S1S1P0 prism has a refractive index of 1.64 for blue light and 1.61 for red light.Calculate the difference in the angle of incidence for the initial red and blue light ray that is refracted inside the prism and just undergoes total internal reflection at the opposite face of the prism.
A)4.1S1U1P1$\circ$S1S1P0
B)1.5S1U1P1$\circ$S1S1P0
C)3.2S1U1P1$\circ$S1S1P0
D)2.4S1U1P1$\circ$S1S1P0
E)0.4S1U1P1$\circ$S1S1P0

Accepted Answer

The critical angle for total internal reflection is different for blue and red light due to their different refractive indices. The difference in the angle of incidence is calculated using Snell's law and the critical angle formula for each color.

Quiz 31: Properties of Light

The wavelength of green light in air is 500 nm.The wavelength of this light in glass of index of refraction 1.60 is approximately

A light wave traveling in air impinges on an amber plate at an angle of incidence of 60º.If the angle of refraction is 34º,the velocity of light in amber must be approximately

A small beam of monochromatic light in air is incident on a large plate of glass with plane-parallel sides at an angle of 30º.The index of refraction of the glass plate for this wavelength is 1.50.The angle of emergence of the light on the far side of the plate is approximately

Two clear but non-mixing liquids each of depth 15 cm are placed together in a glass container.The liquids have refractive indices of 1.75 and 1.33.What is the apparent depth of the combination when viewed from above?

A 60^$\circ$ prism has a refractive index of 1.64 for blue light and 1.61 for red light.Calculate the difference in the angle of emergence from the opposite face of the prism for a beam of red and a beam of blue light that are both incident at an angle of 60^o.

A 60^$\circ$ prism has a refractive index of 1.52.Calculate the angle of incidence for the initial light ray that is then refracted inside the prism and just undergoes total internal reflection at the opposite face of the prism.

A light wave in air strikes a glass plate 1 cm thick at an angle of incidence of 50.0º.If the index of refraction of the glass is 1.50 for this light,the actual path length of the light through the glass is approximately

If the refractive index of glass is 1.5 and the refractive index of a particular liquid is 1.38,then calculate the value for the critical angle of total internal reflection at the liquid-glass interface.

You shine light of frequency 4.00 $\times$ 10¹⁴Hz through a glass plate whose index of refraction for this light is 1.50.The wavelength of this light in the glass is approximately

The minimum angle of deviation produced by a glass prism (n = 1.70 for the light used) whose apex angle is 45º is approximately

A column of liquid is 10 cm high.Viewed from above the liquid has an apparent depth of 6.77 cm.If the refractive index of the liquid is 1.33 at the top and increases uniformly to the bottom,then what is the refractive index at the bottom of the liquid?

A 60^$\circ$ prism has a refractive index of 1.71.Calculate the angle of emergence out of the prism for a ray that enters one side at an angle almost parallel to one face,i.e.at 90^$\circ$ to the normal.

The critical angle for light rays passing from flint glass (n = 1.70) into air is about

A 60^$\circ$ prism has a refractive index of 1.64 for blue light and 1.61 for red light.Calculate the difference in the angle of incidence for the initial red and blue light ray that is refracted inside the prism and just undergoes total internal reflection at the opposite face of the prism.