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konstantin123 [22]

3 years ago

6

In a Young's double-slit experiment, a set of parallel slits with a separation of 0.114 mm is illuminated by light having a wave

length of 558 nm and the interference pattern observed on a screen 4.50 m from the slits. (a) What is the difference in path lengths from the two slits to the location of a fourth order bright fringe on the screen? μm (b) What is the difference in path lengths from the two slits to the location of the fourth dark fringe on the screen, away from the center of the pattern?

1 answer:

elena-14-01-66 [18.8K]3 years ago

7 0

Answer:

a) 2232nm

b) 2511nm

Explanation:

a) To find the path difference Δl you use the following formula:

$\Delta l=m\lambda$

m: order of a bright fringe

λ: wavelength of light = 558nm

for m=4:

$\Delta l=(4)(558nm)=2232nm$

b) The path difference for the case of destructive interference you have:

$\Delta l=(m+\frac{1}{2})\lambda$

m: order of a dark fringe

for m=4:

$\Delta l=(4+\frac{1}{2})(558nm)=2511nm$

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Match the organisms to the descriptions.

makvit [3.9K]

Answer:

Lacks colored blood - Scorpion

Soft, unsegmented body - Octopus

A vertebrate - Bird

Lacks antennae - Starfish

Explanation:

Scorpion belongs to the Phylum Arthropoda of the Kingdom Animalia. All the organisms of this phylum lack coloured blood.

Octopus comes under the Phylum Molusca of the Kingdom Animalia. Soft, unsegmented body is the property of the organisms of this phylum.

Bird belongs to the Vertebrata class of the Phylum Chordata of Kingdom Animalia.

Starfish belongs to Echinodermata Phylum of the Kingdom Animalia. The organisms of this category do not posses antennae.

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4 years ago

Read 2 more answers

Describe the process of sexual reproduction in terms of the life cycle of cells.

umka2103 [35]

The first growth phase (G1): During the G1 stage, the cell doubles in size and doubles the number of organelles.

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5 0

3 years ago

A 2.81 μF capacitor is charged to 1220 V and a 6.61 μF capacitor is charged to 560 V. These capacitors are then disconnected fro

fenix001 [56]

Answer:

756.88 Volts will be the potential difference across each capacitor.

Explanation:

$Q=C\times V$

Q = Charge on capacitor

C = Capacitance

V = Voltage across capacitor

Capacitance of first capacitor = $C_1=2.81 \mu F=2.81\times 10^{-6} F$

Charge of first capacitor = $Q_1$

Voltage across first capacitor = $V_1=1220 V$

$Q_1=C_1V_1$

$Q_1=2.81\times 10^{-6} F\times 1220 V=0.0034282 C$

Capacitance of first capacitor = $C_2=6.61\mu F=6.61\times 10^{-6} F$

Charge of second capacitor = $Q_2$

Voltage across first capacitor = $V_2=560 V$

$Q_2=C_2V_2$

$Q_1=6.61\times 10^{-6} F\times 560 V=0.0037016 C$

Both the capacitors are disconnected and positive plates are now connected to each other and the negative plates are connected to each other. These capacitors are connected in parallel combination.

Total charge = Q

$Q =Q_1+Q_2=0.0034282 C+ 0.0037016 C=0.0071298 C$

Total capacitance in parallel combination:

$C_p=C_1+C_2=2.81\times 10^{-6} F+6.61\times 10^{-6} F=9.42\times 10^{-6} F$

Potential across both capacitors = V

$V=\frac{Q}{C}=\frac{0.0071298 C}{9.42\times 10^{-6} F}=756.88 V$

756.88 Volts will be the potential difference across each capacitor.

8 0

3 years ago

1. What are the three ways an object can accelerate?

solong [7]

Answer:

increase speed, decrease speed, and change direction

Explanation:

4 0

3 years ago

A skateboarder is moving at 1.75 m/s when she starts going up an incline that causes an acceleration of -0.20 m/s2

Rudiy27

Answer:

Approximately $7.66\; \rm m$ .

Explanation:

<h3>Solve this question with a speed-time plot</h3>

The skateboarder started with an initial speed of $u = 1.75\; \rm m \cdot s^{-1}$ and came to a stop when her speed became $v = 0\; \rm m \cdot s^{-1}$ . How much time would that take if her acceleration is $a = -0.20\; \rm m \cdot s^{-1}$ ?

$\begin{aligned} t &= \frac{v - u}{a} \\ &= \frac{0\; \rm m \cdot s^{-1} - 1.75\; \rm m \cdot s^{-1}}{-0.20\; \rm m \cdot s^{-2}} \approx 8.75\; \rm s\end{aligned}$ .

Refer to the speed-time graph in the diagram attached. This diagram shows the velocity-time plot of this skateboarder between the time she reached the incline and the time when she came to a stop. This plot, along with the vertical speed axis and the horizontal time axis, form a triangle. The area of this triangle should be equal to the distance that the skateboarder travelled while she was moving up this incline until she came to a stop. For this particular question, that area is approximately equal to:

$\displaystyle \frac{1}{2} \times 1.75\; \rm m \cdot s^{-1} \times 8.75\; \rm s \approx 7.66\; \rm m$ .

In other words, the skateboarder travelled $15.3\; \rm m$ up the slope until she came to a stop.

<h3>Solve this question with an SUVAT equation</h3>

A more general equation for this kind of motion is:

$\displaystyle x = \frac{1}{2}\, (u + v) \, t = \frac{1}{2}\, (u + v)\cdot \frac{v - u}{a}= \frac{v^2 - u^2}{2\, a}$ ,

where:

$u$ and $v$ are the initial and final velocity of the object,
$a$ is the constant acceleration that changed the velocity of this object from $u$ to $v$ , and
$x$ is the distance that this object travelled while its velocity changed from $u$ to $v$ .

For the skateboarder in this question:

$\begin{aligned}x &= \frac{v^2 - u^2}{2\, a}\\ &= \frac{\left(0\; \rm m \cdot s^{-1}\right)^2 - \left(1.75\; \rm m \cdot s^{-1}\right)^2}{2\times \left(-0.20\; \rm m \cdot s^{-2}\right)}\approx 7.66\; \rm m \end{aligned}$ .

6 0

4 years ago