Ask question

Ask question

All categories

2 years ago

12

For a demonstration, a professor uses a razor blade to cut a thin slit in a piece of aluminum foil. When she shines a laser poin

ter (λ=680nm) through the slit onto a screen 5.4 m away, a diffraction pattern appears. The bright band in the center of the pattern is 7.9 cm wide. What is the width of slide?

1 answer:

nydimaria [60]2 years ago

4 0

Answer:

The width of slide is 0.092 mm.

Explanation:

Given that,

Wave length = 680 nm

Distance between slit and screen D= 5.4 m

Distance of bright band 2y = 7.9 cm

$y =\dfrac{7.9}{2}$

$y=3.95\ cm$

We need to calculate the width of slide

Using formula of width

$d=\dfrac{\lambda D}{y}$

Where, d = width

D =Distance between slit and screen

y = Distance of bright band

Put the value into the formula

$d=\dfrac{680\times10^{-9}\times5.4}{3.95\times10^{-2}}$

$d=0.092\times10^{-3}\ m$

$d=0.092\ mm$

Hence, The width of slide is 0.092 mm.

You might be interested in

The eyes of amphibians such as frogs have a much flatter cornea but a more strongly curved (almost spherical) lens than do the e

Lapatulllka [165]

Answer:

0.2cm towards the retina.

Explanation:

the focal length of the frog eye is

(1/f) = (1/10) + (1/0.8)

f = 0.74cm

Since the distance of the object is 15cm Hence

(1/0.74) = (1/15) + (1/V)

V = 0.78cm

Therefore the distance the retina is to move is

0.78cm - 0.8cm = 0.02cm towards the retina.

3 0

2 years ago

You are piloting a helicopter which is rising vertically at a uniform velocity of 14.70 m/s. When you reach 196.00 m, you see Ba

Cloud [144]

Answer:

The ball reaches Barney head in $t = 8 \ s$

Explanation:

From the question we are told that

The rise velocity is $v = 14.70 \ m/s$

The height considered is $h = 196 \ m$

The horizontal velocity of the large object is $v_h = 8.50 \ m/s$

Generally from kinematic equation

$s = ut + \frac{1}{2} gt^2$

Here s is the distance of the object from Barney head ,

u is the velocity of the object along the vertical axis which is equal but opposite to the velocity of the helicopter

So

$u = -14.7 m/s$

So

$196 = -14.7 t + \frac{1}{2} * 9.8 * t^2$

= $4.9 t^2 - 14.7t - 196 = 0$

Solving the above equation using quadratic formula

The value of t obtained is $t = 8 \ s$

6 0

2 years ago

A seaplane flies horizontally over the ocean at 50 meters/second. It releases a buoy, which lands after 21 seconds. What's the v

pantera1 [17]

The motion of the buoy consists of two independent motions on the horizontal and vertical axis.

On the horizontal axis, the motion of the buoy is a uniform motion with constant speed $v=50 m/s$ . On the vertical axis, the motion of the buoy is a uniformly accelerated motion with constant acceleration $g=9.81 m/s^2$ . The vertical position of the buoy at time t is given by
$y(t)=h- \frac{1}{2}gt^2$
where h is the initial heigth of the buoy when it is released from the plane. At the time t=21 s, the buoy reaches the ground, so y(21 s)=0. If we substitute these two numbers inside the equation, we can find the value of h, the vertical displacement from the plane to the ocean:
$0=h- \frac{1}{2}gt^2$
$h= \frac{1}{2}gt^2= \frac{1}{2}(9.81 m/s^2)(21 s)^2=2163 m$

8 0

2 years ago

A beam of light has a wavelength of 4.5 x10^-7 meter in a vacuum. the frequency of this light is

valkas [14]

The basic relationship between frequency and wavelength for light (which is an electromagnetic wave) is
$c= f \lambda$
where c is the speed of light, f the frequency and $\lambda$ the wavelength of the wave.
Using $\lambda=4.5 \cdot 10^{-7} m$ and $c=3 \cdot 10^8 m/s$ , we can find the value of the frequency:
$f= \frac{c}{\lambda}= \frac{3 \cdot 10^8 m/s}{4.5 \cdot 10^{-7} m}=6.7 \cdot 10^{14} Hz$

3 0

2 years ago

A rigid tank whose volume is unknown is divided into two parts by a partition. One side of the tank contains an ideal gas at 935

a_sh-v [17]

Answer:

2805 °C

Explanation:

If the gas in the tank behaves as ideal gas at the start and end of the process. We can use the following equation:

$P=RTn/V$

The key issue is identify the quantities (P,T, V, n) in the initial and final state, particularly the quantities that change.

In the initial situation the gas have an initial volume $V_{i}$ , temperature $T_{i}$ , and pressure $P$ ,.

And in the final situation the gas have different volume $V_{f}$ and temeperature $T_{f}$ , the same pressure $P$ ,, and the same number of moles $n$ ,.

We can write the gas ideal equation for each state:

$P=RT_{i}n/V_{i}$ and $P=RT_{f}n/V_{f}$ , as the pressure are equals in both states we can write

$RT_{i}n/V_{i} = RT_{f}n/V_{f}$

solving for $T_{f}$

$T_{f} = T_{i}/V_{i} * V_{f}$ (*)

We know $T_{i}$ = 935 °C, and that the $V_{f}$ (the complete volume of the tank) is the initial volume $V_{i}$ plus the part initially without gas which has a volume twice the size of the initial volume (read in the statement: the other side has a volume twice the size of the part containing the gas). So the final volume $V_{f}= V_{i} + 2V_{i}=3V_{i}$

Replacing in (*)

$T_{f} = 935/V_{i} * 3V_{i} = 935*3= 2805$

7 0

1 year ago