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kifflom [539]
2 years ago
8

After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of r

efraction of 1.38, while the eyedrops have a refractive index of 1.45. After you put in the drops, your friends notice that your eyes look red, because red light of wavelength 600 nm has been reinforced in the reflected light.Part A What is the minimum thickness of the film of eyedrops on your cornea?t= nm.
Physics
1 answer:
mina [271]2 years ago
6 0

Answer:

t = 103.45 n m

Explanation:

given,

refractive index of cornea = 1.38

refractive index of eye drop = 1.45

wavelength of refractive index = 600 nm

refractive index of eye drop is greater than refractive index of cornea and the air.

Formula used in this case

for constructive interference

2 n t = (m + \dfrac{1}{2})\lambda

At m = 0 for the minimum thickness, so

2\times 1.45 \times t = (0 + 0.5)\times 600

2.9 \times t =300

t =\dfrac{300}{2.9}

t = 103.45 n m

the minimum thickness of the film of eyedrops t = 103.45 n m

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harkovskaia [24]

Answer:

λ = 2042 nm

Explanation:

given data

screen distance d = 11 m

spot s = 4.5 cm = 4.5 ×10^{-2} m

separation L = 0.5 mm = 0.5 ×10^{-3} m

to find out

what is λ

solution

we will find first angle between first max and central bright

that is tan θ = s/d

tan θ = 4.5 ×10^{-2}  / 11

θ = 0.234

and we know diffraction grating for max

L sinθ  = mλ

here we know m = 1  so put all value and find λ

L sinθ  = mλ

0.5 ×10^{-3}  sin(0.234)  = 1 λ

λ = 2042.02 ×10^{-9}  m

λ = 2042 nm

3 0
2 years ago
A 35 g steel ball is held by a ceiling-mounted electromagnet 4.0 m above the floor. A compressed-air cannon sits on the floor, 4
HACTEHA [7]

Answer:

7.9 m/s

Explanation:

When both balls collide, they have spent the same time for their motions.

Motion of steel ball

This is purely under gravity. It is vertical.

Initial velocity, <em>u </em>= 0 m/s

Distance, <em>s</em> = 4.0 m - 1.2 m = 2.8 m

Acceleration, <em>a</em> = g

Using the equation of motion

s = ut+\frac{1}{2}at^2

2.8 \text{ m} = 0+\dfrac{gt^2}{2}

t = \sqrt{\dfrac{5.6}{g}}

Motion of plastic ball

This has two components: a vertical and a horizontal.

The vertical motion is under gravity.

Considering the vertical motion,

Initial velocity, <em>u </em>= ?

Distance, <em>s</em> = 1.2 m

Acceleration, <em>a</em> = -<em>g                   </em> (It is going up)

Using the equation of motion

s = ut+\frac{1}{2}at^2

1.2\text{ m} = ut-\frac{1}{2}gt^2

Substituting the value of <em>t</em> from the previous equation,

1.2\text{ m} = u\sqrt{\dfrac{5.6}{g}}-\dfrac{1}{2}\times g\times\dfrac{5.6}{g}

u\sqrt{\dfrac{5.6}{g}} = 4.0

Taking <em>g</em> = 9.8 m/s²,

u = \dfrac{4.0}{0.756} = 5.29 \text{ m/s}

This is the vertical component of the initial velocity

Considering the horizontal motion which is not accelerated,

horizontal component of the initial velocity is horizontal distance ÷ time.

u_h = \dfrac{4.4\text{ m}}{0.756\text{ s}} = 5.82\text{ m/s}

The initial velocity is

v_i = \sqrt{u^2+u_h^2} = \sqrt{(5.29\text{ m/s})^2+(5.82\text{ m/s})^2} = 7.9 \text{ m/s}

4 0
2 years ago
Consider an alcohol and a mercury thermometer that read exactly 0 oC at the ice point and 100 oC at the steam point. The distanc
Alexeev081 [22]

Answer:

No, both the thermometers will give the different reading.

Explanation:

Given,

  • Both thermometer has same ice point = T_i\ =\ 0^o C
  • Both thermometer has same steam point = T_s\ =\ 100^o C
  • Distance between the ice point and steam point in both the thermometer is same of 100 division,

All the data given in both the thermometers are same, but the material in the thermometer is different due to this the reading at 60^o C will differ in both the thermometer. Because the reading on both the thermometer is depended upon the thermal expansion of the material inside it, but both the materials are different. Due to this the rise of fluid in the thermometer, i,e,. the volume of the fluid material in the thermometer will depend upon the thermal expansion.  Hence both the material alcohol and mercury have the different thermal expansion, therefore the rise of the fluid in the thermometer also differ in both the thermometer.

7 0
2 years ago
A block spring system oscillates on a frictionless surface with an amplitude of 10\text{ cm}10 cm and has an energy of 2.5 \text
antoniya [11.8K]

Answer:

The energy of the system is 15 J.

Explanation:

Given that,

Energy E = 2.5 J

Amplitude = 10 cm

We need to calculate the spring constant

Using formula of mechanical energy of the system

E=\dfrac{1}{2}kA^2

Put the value into the formula

2.5=\dfrac{1}{2}k\times(10\times10^{-2})^2

k=\dfrac{2.5\times2}{(10\times10^{-2})^2}

k=500\ N/m

If the block is replaced by a block with twice the mass of the original block

Amplitude = 6 cm

We need to calculate the energy

Using formula of mechanical energy

E=\dfrac{1}{2}kA^2

Put the value into the formula

E=\dfrac{1}{2}\times500\times(6\times10^{-2})

E=15\ J

Hence, The energy of the system is 15 J.

8 0
2 years ago
A pulley 12 cm in diameter is free to rotate about a horizontal axle. A 220-g mass and a 470-g mass are tied to either end of a
Aleks [24]

Answer:0.147 N-m

Explanation:

Given

Diameter of Pulley d=12 cm

radius r=6 cm

mass of first object m_1=220 gm

mass of second object m_2=470 gm

Now both masses will exert a torque a on Pulley

Torque due to first Pulley T_1=m_1g\cdot r

T_1=0.22\times 9.8\times 0.06=0.129 N-m

Torque due to second mass on Pulley T_2=m_2g\cdot r

T_2=0.276 N-m

Total Torque by masses T_{net}=T_2-T_1

T_{net}=0.276-0.129=0.147 N-m

so we need to apply a torque of magnitude 0.147 N-m opposite to the direction of T_{net}  

3 0
2 years ago
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