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kvv77 [185]
3 years ago
11

Find the separation of two points on the Moon’s surface that can just be resolved by the 200 in. (= 5.1 m) telescope at Mount Pa

lomar, assuming that this separation is determined by diffraction effects. The distance from Earth to the Moon is
3.8×105km
. Assume a wavelength of 550 nm for the light.
Physics
1 answer:
rewona [7]3 years ago
6 0

Answer:

The separation of the 2 points should be 50.0 meters.

Explanation:

According to Rayleigh's scattering criteria the angular separation between 2 points to be resolved equals

\theta =1.22\cdot \frac{\lambda }{D}

Applying the given values we get

\theta =1.22\cdot \frac{550\times 10^{-9} }{5.1}=0.1316\times 10^{-6}rads

thus the linear separation equals L=\theta \times Distance

Applying the given values we get

L=0.1316\times 10^{-6} \times 3.8\times 10^{5}\times 10^{3}meters\\\\\therefore L=50.00metes

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What is the force per unit area at this point acting normal to the surface with unit nor- Side View √√ mal vector n = (1/ 2)ex +
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Complete Question:

Given \sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right] at a point. What is the force per unit area at this point acting normal to the surface with\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z   ? Are there any shear stresses acting on this surface?

Answer:

Force per unit area, \sigma_n = 28 MPa

There are shear stresses acting on the surface since \tau \neq 0

Explanation:

\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]

equation of the normal, \b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z

\b n = \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]

Traction vector on n, T_n = \sigma \b n

T_n =  \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right] \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]

T_n = \left[\begin{array}{ccc}\frac{23}{\sqrt{2} }\\0\\\frac{27}{\sqrt{33} }\end{array}\right]

T_n = \frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z

To get the Force per unit area acting normal to the surface, find the dot product of the traction vector and the normal.

\sigma_n = T_n . \b n

\sigma \b n = (\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z) . ((1/ \sqrt{2} ) \b e_x + 0 \b  e_y +(1/ \sqrt{2}) \b e_z)\\\\\sigma \b n = 28 MPa

If the shear stress, \tau, is calculated and it is not equal to zero, this means there are shear stresses.

\tau = T_n  - \sigma_n \b n

\tau =  [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - 28( (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z)\\\\\tau =  [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - [ (28/ \sqrt{2} ) \b e_x + (28/ \sqrt{2}) \b e_z]\\\\\tau =  \frac{-5}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{5}{\sqrt{2} } \b e_z

\tau = \sqrt{(-5/\sqrt{2})^2  + (27/\sqrt{2})^2 + (5/\sqrt{2})^2} \\\\ \tau = 19.74 MPa

Since \tau \neq 0, there are shear stresses acting on the surface.

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