Ask question

Ask question

All categories

2 years ago

9

The radius of curvature of both sides of a converging lens is 18 cm. One side of the lens is coated withsilver so that the inner

surface is reflective. When light is incident on the uncoated side it passes throughthe lens, reflects off the silver coating, and passes back through the lens. The overall effect is that of amirror with focal length 5.0 cm. What is the index of refraction of the lens material

1 answer:

Dahasolnce [82]2 years ago

7 0

Answer:

n = 1.4

Explanation:

Given,

R1 = 18 cm, R2 = -18 cm

From lens makers formula

1/f = (n - 1)(1/18 + 1/18) = (n-1)/9

f = 9/(n-1)

Power, P = 1/f ( in m) = (n-1)/0.09

Now, this lens is in with conjunction with a concave mirror which then can be thought of as to be in conjunction with another thin lens

Power of concave mirror = P' = 1/f ( in m) = 2/R = 2/0.18 = 1/0.09

Net power of the combination = 2P + P' = 2(n-1)/0.09 + 1/0.09 = 1/0.05

n = 1.4

You might be interested in

It's a frightening idea, but what would be the sound intensity level of 100 physics professors talking simultaneously

Svetlanka [38]

The sound intensity level of 100 physics professors talking simultaneously is equal to 72 decibel.

<u>Given the following data:</u>

Sound intensity = 52 dB.

<h3>What is sound intensity?</h3>

Sound intensity can be defined as a measure of the power of a sound wave per unit area. Thus, sound intensity is a quantity that can be used to measure the energy of sound and its unit of measurement is Watts per square meter.

<h3>How to calculate the sound intensity level.</h3>

Mathematically, sound intensity level is given by this formula:

$\beta = 10 log(\frac{I}{I_{ref}} )$

<u>Note:</u> The reference value of sound intensity is equal to $1.0 \times 10^{-12}\;W/m^2$

Thus, the sound intensity for one (1) professor is given by:

$I_1 = 1.0 \times 10^{-12} \times 10^{5.2}\\\\I_1=1.585 \times 10^{-7}\;W/m^2$

For 100 professors, the sound intensity is:

$I_{100} = 1.585 \times 10^{-7} \times 100\\\\I_{100}=1.585 \times 10^{-5}\;W/m^2$

Substituting the parameters into the formula, we have;

$\beta = 10 log(\frac{1.585 \times 10^{-5}}{1.0 \times 10^{-12}})\\\\\beta = 10 log(15.58 \times 10^{6})\\\\\beta = 10 \times 7.2\\\\\beta =72\;dB$

Read more on sound intensity here: brainly.com/question/17062836

<u>Complete Question:</u>

One physics professor talking produces a sound intensity level of 52 dB. It's a frightening idea, but what would be the sound intensity level of 100 physics professors talking simultaneously?

7 0

2 years ago

An end of a light wire rod is bent into a hoop of radius r. The straight part of the rod has length l; a ball of mass M is attac

Roman55 [17]

Answer:

$arcsin(\frac{R\mu}{(R+l)\sqrt{\mu^2+1}})$

Explanation:

By the Law of Sines,

$sin \theta = \frac{sin \phi R}{ l + R}$

From Newton's Law,

$mg = N\sqrt{\mu^2+1}$

And the last equation again from Newton's Law,

$\mu N = mgsin\phi$

Then if we collect all equations together,

$\mu N = mgsin\phi = N\sqrt{\mu^2+1}sin\phi\\$

$sin\theta = \frac{\mu R}{ (l + R)\sqrt{\mu^2+1}}$

Thus,

$\theta = arcsin(\frac{R\mu}{(R+l)\sqrt{\mu^2+1}})$

8 0

3 years ago

What an object is placed 8 mm from a concave spherical mirror a clear image can be projected on the screen 16 mm in front of me

alexgriva [62]

Concept: The magnification of spherical mirror can be defined by two ways.

(i) In terms of the height of the object and image.

The magnification of the spherical mirror is defined as the ratio of the height of the image' $h_{i}$ ' to the height of the object ' $h_{o}$ '. It is denoted by letter 'm'.

Mathematically, it can be written as

$m= \frac{h_{i}}{h_{o}} ------------(1)$

(ii) In terms of the object's and image's distances.

The magnification of the spherical mirror is defined as the negative ratio of the image distance' $d_{i}$ ' to the object distance ' $d_{o}$ '.

Mathematically, it can be written as

$m= - \frac{d_{i}}{d_{o}} ------------(2)$

Now, from equation (1) and (2) we have,

$m = \frac{h_{i}}{h_{o}} = - \frac{d_{i}}{d_{o}} -----------(3)$

Given: Spherical Concave Mirror,

We will consider positive sign for object's and image's distance because both are in front of the mirror.

Object distance $(d_{o}) = + 8 mm$ .

Image distance $(d_{i}) = + 16 mm$

Object's height $(h_{o}) = + 4 mm$

Image's height $(h_{i}) =?$

Now, apply equation (3)

$\frac{h_{i}}{h_{o}} = - \frac{d_{i}}{d_{o}}$

$Or, \frac{h_{i}}{4 mm} = - \frac{+16 mm}{+8 mm}$

Or, hi = - 8 mm

Here; negative sign means, the image will be inverted.

The image's height will be 8 mm.

4 0

3 years ago

Explain why a balloon filled with air cannot float in the middle of a room without moving downwards but a helium balloon of the

ELEN [110]

Helium is less dense than normal air

6 0

3 years ago

An object of mass 3.00 kg, moving with an initial velocity of 5.05 m/s, collides with and sticks to an object of mass 2.76 kg wi

Lynna [10]

Answer:

0.752 m/s

Explanation:

m1 = 3.00kg

u1 = 5.05m/s

m2 = 2.76kg

u2 = -3.66m/s

According to the law of conservation of momentum,

m1u1 + m2u2 = (m1+m2)v

3(5.05) + 2.76(-3.66) = (5.05+2.76)v

15.15 - 9.2736 = 7.81v

5.8764 = 7.81v

v = 5.8764/7.81

v = 0.752m/s

6 0

3 years ago