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Marat540 [252]
3 years ago
13

The nonreflective coating on a camera lens with an index of refraction of 1.21 is designed to minimize the reflection of 570-nm

light. If the lens glass has an index of refraction of 1.52, what is the minimum thickness of the coating that will accomplish this ta
Physics
1 answer:
lord [1]3 years ago
5 0

Answer: 117.8 nm

Explanation:

Given,

Nonreflective coating refractive index : n = 1.21

Index of refraction: n_0 = 1.52

Wave length of light = λ = 570 nm = 570\times10^{-9}\ m

\text{ Thickness}=\dfrac{\lambda}{4n}

=\dfrac{570\times10^{-9}\ m}{4\times1.21}\\\\\approx\dfrac{117.8\times 10^{-9}\ m}{1}\\\\=117.8\text{ nm}

Hence, the minimum thickness of the coating that will accomplish= 117.8 nm

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Help wit these questions someone.
romanna [79]

In series circuit, Req = R₁ + R₂ + R₃ + ···

In parallel circuit, \frac{1}{Req}  = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3} +...

<h3>Q7.</h3>

total resistance in the upper branch = R₂ + R₃ = R₂ + 2

\frac{1}{Req} = \frac{1}{R2+R3} + \frac{1}{R1}

\frac{1}{4} = \frac{1}{R2+2} +\frac{1}{6}

R₂ + 2 = 12

R₂ = 10Ω

<h3>Q8.</h3>

\frac{1}{Req} = \frac{1}{R2+R3} + \frac{1}{R1}

\frac{1}{Req} = \frac{1}{2+1} + \frac{1}{4}

Req = 1.7Ω

7 0
3 years ago
What is the correct path of sperm cells through the male reproductive system?
irakobra [83]

Answer:

What is the correct path of sperm cells through the male reproductive system?

Epididymis, seminiferous tubules, urethra, vas deferens

<u>Seminiferous tubules, epididymis, vas deferens, urethra </u>

Urethra, seminiferous tubules, epididymis, vas deferens

Seminiferous tubules, vas deferens, epididymis, urethra

Hope this helps :)

Have a great day !

5INGH

Explanation:

8 0
2 years ago
A bird flies directly into the windshield of a moving car. How does the FORCE exerted on the car compare to the FORCE exerted on
Anarel [89]

Answer:

b

Explanation:

7 0
3 years ago
A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the
BartSMP [9]

Answer:

r=1.14m

Explanation:

\theta is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

F_m=F_c\\qvB=F_c(1)

F_c is the centripetal force and is defined as:

F_c=m\frac{v^2}{r}

Here v is the proton's speed and r is the radius of the circular motion. Replacing this in (1) and solving for r:

qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}

Recall that 1 J is equal to 6.242*10^{12}MeV, so:

4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J

We can calculate v from the kinetic energy of the proton:

K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}

Finally, we calculate the radius of the proton path:

r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m

8 0
3 years ago
Que es la expansión del universo?
anastassius [24]

Answer:

La expansión no es más que el incremento con el tiempo de la distancia entre cualquier par de galaxias lejanas. Se suele utilizar para representar este hecho la analogía de un globo donde hemos pintado una serie de puntos a modo de galaxias.

Explanation:

3 0
2 years ago
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