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garri49 [273]
3 years ago
8

Chromatic aberration comes from the fact that different wavelengths of light travel at different speeds through the material of

the lens; that is, they have different indices of refraction, a property known as dispersion. This means that a lens, in effect, has different focal lengths for different wavelengths of light. Consider a lens made to the following specifications: focal length for red light fred=19.57cm, focal length for blue light fblue=18.87cm.
Part A

Consider an object 5.000cm tall placed a distance 30.00cm from the lens. Assuming that the object reflects both red and blue light, find the ratio yredyblueof the height of the red image to the height of the blue image.

Express your answer numerically to three significant figures.
Physics
1 answer:
gtnhenbr [62]3 years ago
5 0

Answer:

 y_red / y_blue = 1.11

Explanation:

Let's use the constructor equation to find the image for each wavelength

         1 /f = 1 /o + 1 /i

Where f is the focal length, or the distance to the object and i the distance to the image

Red light

           1 / i = 1 / f - 1 / o

           1 / i_red = 1 / f_red - 1 / o

           1 / i_red = 1 / 19.57 - 1/30

           1 / i_red = 1,776 10-2

           i_red = 56.29 cm

Blue light

            1 / i_blue = 1 / f_blue - 1 / o

            1 / i_blue = 1 / 18.87 - 1/30

            1 / i_blue = 1,966 10-2

            i_blue = 50.863 cm

Now let's use the magnification ratio

             m = y ’/ h = - i / o

             y ’= - h i / o

Red Light

            y_red ’= - 5 56.29 / 30

            y_red ’= - 9.3816 cm

Light blue

            y_blue ’= 5 50,863 / 30

            y_blue ’= - 8.47716 cm

The ratio of the height of the two images is

            y_red ’/ y_blue’ = 9.3816 / 8.47716

            y_red / y_blue = 1,107

            y_red / y_blue = 1.11

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Tranverse Waves

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1 year ago
Two electrons are at rest and separated by a distance of 4.32 × 10-10 m. When they are released they accelerate away from each o
sasho [114]

Answer:

Speed of electron when their separation increased by a factor of 4.10 is 9.41 x 10⁵ m/s .

Explanation:

The electric potential energy is given by the relation :

U = \frac{kq_{1}q_{2}  }{r}

Here q₁ and q₂ are the two charge particles and r is the distance between them and k is electric constant.

In this case, there are two electrons which are separated by the distance 4.32 x 10⁻¹⁰ m.

Let e be the electron charge and r₁ be the distance between them. Then, the initial electric potential energy is :

U_{1}  = \frac{ke^{2}   }{r_{1} }

Now, the distance between the electrons increases by the factor of 4.10. Let r₂ be the new distance between them i.e. r₂ = 4.10 r₁.

Thus, the new electric potential energy is :

U_{2}  = \frac{ke^{2}   }{r_{2} }=\frac{ke^{2}   }{4.10r_{1} }

Applying law of conservation of energy :

ΔU  = ΔK

Here ΔU is change in electric potential energy and ΔK is change in kinetic energy.

( U₁  - U₂ ) = ( K₂ - K₁ )

Here K₂ and K₁ are initial and final kinetic energy of electron.

Since, the electron initially is at rest, so its initial kinetic energy is zero. Thus, the above equation becomes:

K₂ = U₁ - U₂

\frac{1}{2}mv^{2}=\frac{ke^{2}   }{r_{1} }- \frac{ke^{2}   }{4.10r_{1} }

Here m and v are the mass and final speed of electron respectively.

v^{2}=\frac{2}{m} \frac{ke^{2}   }{r_{1} }(1- \frac{1  }{4.10 })

Substitute 9.1 x 10⁻³¹ kg for m, 9 x 10⁹ N m² C⁻² for k, 1.6 x 10⁻¹⁹ C for e and 4.32 x 10⁻¹⁰ m for r₁ in the above equation.

v^{2}=\frac{2}{9.1\times10^{-31} } \frac{9\times10^{9}\times(1.6\times10^{-19})^{2}   }{4.32\times10^{-10} }(1- \frac{1  }{4.10 })

v^{2}=8.86\times10^{11}

v = 9.41 x 10⁵ m/s

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A girl pushes a 1.04 kg book across a table with a horizontal applied force 10 points
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Answer:

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As a side note, the F_N and W on this book are not equal- these two forces are equal in size but point in the opposite directions.

When the book is moving, the friction F(\text{kinetic friction}) on it will be equal to

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\begin{aligned}& F(\text{kinetic friction}) \\ &= \mu_{\rm k}\cdot F(\text{normal force})\\ &\approx 0.35 \times 10.202\; \rm N \approx 3.5708\; \rm N\end{aligned}.

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\begin{aligned}& F(\text{net force})  \\ &= 15.0 \; \rm N - F(\text{kinetic friction}) \\& \approx 15.0 - 3.5708\; \rm N \approx 11.429\; \rm N\end{aligned}.

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\displaystyle a = \frac{F(\text{net force})}{m} \approx \frac{11.429\; \rm N}{1.04\; \rm kg} \approx 11.0\; \rm m \cdot s^{-2}.

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