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3 years ago

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(a) Using the Rayleigh criterion, estimate the theoretical diffraction limit for the angular resolution of a typical 20-cm (8-in

) amateur telescope at 550 nm. Express your answer in arcseconds.

1 answer:

Andreas93 [3]3 years ago

8 0

Answer:

0.692" arc sec.

Explanation:

Tan Ꝋ = Sin Ꝋ = ꝊR = 1.22 (Wavelength) / d

ꝊR = 1.22 x 550 x 10^-9m / 0.2m

= <em>3.355 x 10^-6 rad</em> x (3600 × 180) / π = 0.692" arc sec.

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3 years ago

wo parallel-plate capacitors C1 and C2 are connected in parallel to a 12.0 V battery. Both capacitors have the same plate area o

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Complete Question

Two parallel-plate capacitors C1 and C2 are connected in parallel to a 12.0 V battery. Both capacitors have the same plate area of 5.30 cm2 and plate separation of 2.65 mm. However, the first capacitor C1 is filled with air, while the second capacitor C2 is filled with a dielectric that has a dielectric constant of 2.10.

(a) What is the charge stored on each capacitor

(b) What is the total charge stored in the parallel combination?

Answer:

a

i $Q_1 = 2.124 *10^{-11} \ C$

ii $Q_2 = 4.4604 *10^{-11} \ C$

b

$Q_{eq} = 6.5844 *10^{-11} \ C$

Explanation:

From the question we are told that

The voltage of the battery is $V = 12.0 \ V$

The plate area of each capacitor is $A = 5.30 \ cm^2 = 5.30 *10^{-4} \ m^2$

The separation between the plates is $d = 2.65 \ mm = 2.65 *10^{-3} \ m$

The permittivity of free space has a value $\epsilon_o = 8.85 *10^{-12} \ F/m$

The dielectric constant of the other material is $z = 2.10$

The capacitance of the first capacitor is mathematically represented as

$C_1 = \frac{\epsilon * A }{d }$

substituting values

$C_1 = \frac{8.85 *10^{-12 } * 5.30 *10^{-4} }{2.65 *10^{-3} }$

$C_1 = 1.77 *10^{-12} \ F$

The charge stored in the first capacitor is

$Q_1 = C_1 * V$

substituting values

$Q_1 = 1.77 *10^{-12} * 12$

$Q_1 = 2.124 *10^{-11} \ C$

The capacitance of the second capacitor is mathematically represented as

$C_2 = \frac{ z * \epsilon * A }{d }$

substituting values

$C_1 = \frac{ 2.10 *8.85 *10^{-12 } * 5.30 *10^{-4} }{2.65 *10^{-3} }$

$C_1 = 3.717 *10^{-12} \ F$

The charge stored in the second capacitor is

$Q_2 = C_2 * V$

substituting values

$Q_2 = 3.717*10^{-12} * 12$

$Q_2 = 4.4604 *10^{-11} \ C$

Now the total charge stored in the parallel combination is mathematically represented as

$Q_{eq} = Q_1 + Q_2$

substituting values

$Q_{eq} = 4.4604 *10^{-11} + 2.124*10^{-11}$

$Q_{eq} = 6.5844 *10^{-11} \ C$

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4 years ago