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3 months ago

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Light from a monochromatic source shines through a double slit onto a screen 5.00 m away. The slits are 0.180 mm apart. The dark

bands on the screen are measured to be 1.70 cm apart. What is the wavelength of the incident light? (A) 612 nm (B) 457 nm (C) 306 nm (D) 392 nm

1 answer:

kicyunya [3.2K]3 months ago

4 0

Answer:

The wavelength of the incident light, $\lambda = 612 nm$

Given:

Distance from the slit to the screen, x = 5.00 m

Slit width, d = 0.180 mm

Fringe width, $\beta = 1.70 cm = 0.017 m$

Solution:

To determine the wavelength of the incident light, $\lambda$ :

$\beta = \frac{x\lambda }{d}$

$\lambda = \frac{\beta d}{x}$

$\lambda = \frac{0.017\times 0.180\times 10^{- 3}}{5} = 6.12\times 10^{- 7}m = 612 nm$

$\lambda = 612 nm$

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2 months ago

a. For a spring-mass oscillator, if you double the mass but keep the stiffness the same, by what numerical factor does the perio

kicyunya [3294]

Answer:

a) factor $b=\sqrt{2}$

b) factor $b=\frac{1}{2}$

c) factor $b=1$

d) factor $b=1$

Explanation:

For an oscillating spring-mass system, the time period is expressed as:

$T=\frac{1}{f}$

$T={2\pi} \sqrt{\frac{m}{k} }$

where:

$f=$ represents the frequency of oscillation

$m=$ signifies the mass linked to the spring

$k=$ is the spring's stiffness constant

a) If the mass is doubled:

New mass, $m'=2m$

Thus, the new time period:

$T'=2\pi\sqrt{\frac{m'}{k} }$

$T'=2\pi\sqrt{\frac{2m}{k} }$

$T'=\sqrt{2}\times 2\pi\sqrt{\frac{m}{k} } }$

$T'=\sqrt{2} \times T$

this leads to factor $b=\sqrt{2}$ as per the question.

b) When the stiffness constant is quadrupled, holding other factors constant:

New stiffness constant, $k'=4k$

Thus, the new time period:

$T'=2\pi\sqrt{\frac{m}{k'} }\\\\T'=2\pi\sqrt{\frac{m}{4k} }\\\\T'=\frac{1}{2} \times 2\pi\sqrt{\frac{m}{k} } }\\\\T'=\frac{1}{2} \times T$

this results in factor $b=\frac{1}{2}$ as required.

c) When both mass and stiffness constant are quadrupled:

New stiffness, $k'=4k$

New mass, $m'=4m$

Thus, the new time period:

$T'=2\pi\sqrt{\frac{m'}{k'} }\\\\T'=2\pi\sqrt{\frac{4m}{4k} }\\\\T'=1 \times 2\pi\sqrt{\frac{m}{k} } }\\\\T'=1 \times T$

which leads to factor $b=1$ as stated in the question.

d) If amplitude is quadrupled, the time period remains unaffected because T does not depend on amplitude as demonstrated by the equation.

Thus, factor $b=1$

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