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Ainat [17]
2 years ago
7

A slab of glass 8.0 cm thick is placed upon a printed page. If the refractive index of the glass is 1.5, how far from the surfac

e would the letters appear to be when viewed from directly above?
Physics
1 answer:
7nadin3 [17]2 years ago
8 0

Answer:

5.3 cm

Explanation:

This question is an illustration of real and apparent distance.

From the question, we have the following given parameters

Real Distance, R = 8.0cm

Refractive Index, μ = 1.5

Required

Determine the apparent distance (A)

The relationship between R, A and μ is:

μ = R/A

i.e.

Refractive Index = Real Distance ÷ Apparent Distance

Substitute values in the above formula

1.5 = 8/A

Multiply both sides by A

1.5 * A = A * 8/A

1.5A = 8

Divide both side by 1.5

1.5A/1.5 = 8/1.5

A = 8/1.5

A = 5.3cm

Hence, the letters would appear at a distance of 5.3cm

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3 years ago
A diffraction pattern forms when light passes through a single slit. The wavelength of the light is 691 nm. Determine the angle
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Explanation:

Given that,

Wavelength of the light, \lambda=691\ nm=691\times 10^{-9}\ m

(a) Slit width, a=3.8\times 10^{-4}\ m

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sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-4}}

\theta=0.104^{\circ}

(b) Slit width, a=3.8\times 10^{-6}\ m

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7 0
3 years ago
An electron in the n = 7 level of the hydrogen atom relaxes to a lower energy level, emitting light of 397 nm. what is the value
Dimas [21]

Answer:

n_f=2

Explanation:

It is given that,

Initially, the electron is in n = 7 energy level. When it relaxes to a lower energy level, emitting light of 397 nm. We need to find the value of n for the level to which the electron relaxed. It can be calculate using the formula as :

\dfrac{1}{\lambda}=R(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})

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Solving above equation we get the value of final n is,

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vovikov84 [41]

Answer:

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