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zepelin [54]
3 years ago
11

A. A light wave moves through glass (n=1.5) at an angle of 25°. What angle will it have when it moves from the glass into air (n

=1)?
b. Draw a ray diagram to locate the image of the arrow, as refracted through the lens shown. Write 2 - 3 sentences describing the type of image and its size relative to the object. What type of mirror could be used to form an image of the same type and size? (image attached)


c. An object is located 60 cm from a convex mirror with a focal length of 20 cm. What is the image distance? Is the image real or virtual?

Physics
1 answer:
torisob [31]3 years ago
3 0

PART A)

By Snell's law we know that

n_1sin i = n_2 sin r

here we know that

n_1 = 1.5

i = 25 degree

n_2 = 1

now from above equation we have

1.5 sin25 = 1 sin r

r = 39.3 degree

so it will refract by angle 39.3 degree

PART B)

Here as we can see that image formed on the other side of lens

So it is a real and inverted image

Also we can see  that size of image is lesser than the size of object here

Here we can use concave mirror to form same type of real and inverted image

PART C)

As per the mirror formula we know that

\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}

\frac{1}{d_i} + \frac{1}{60} = \frac{1}{20}

d_i = 30 cm

so image will form at 30 cm from mirror

it is virtual image and smaller in size

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The war between two countries to take over each other's kingdom using weapons to kill each other.

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2 years ago
A hiker is at the bottom of a canyon facing the canyon wall closest to her. She is 790.5 meters from the wall and the sound of h
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Answer:

4.80 seconds

Explanation:

The velocity of sound is obtained from;

V= 2d/t

Where;

V= velocity of sound = 329.2 ms-1

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3 years ago
If your prediction turns out to be false your experiment has been a failure
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3 0
3 years ago
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(a) What is the minimum width of a single slit (in multiples of λ ) that will produce a first minimum for a wavelength λ ? (b) W
Kitty [74]

Answer:

The minimum value of width for first minima is λ

The  minimum value of width for 50 minima is 50λ

The  minimum value of width for 1000 minima is 1000λ

Explanation:

Given that,

Wavelength = λ

For D to be small,

We need to calculate the minimum width

Using formula of minimum width

D\sin\theta=n\lambda

D=\dfrac{n\lambda}{\sin\theta}

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\lambda = wavelength

Put the value into the formula

D=\dfrac{n\lambda}{\sin\theta}

Here, \sin\theta should be maximum.

So. maximum value of \sin\theta is 1

Put the value into the formula

D=\dfrac{1\times\lambda}{1}

D=\lambda

(b). If the minimum number  is 50

Then, the width is

D=\dfrac{50\times\lambda}{1}

D=50\lambda

(c). If the minimum number  is 1000

Then, the width is

D=\dfrac{1000\times\lambda}{1}

D=1000\lambda

Hence, The minimum value of width for first minima is λ

The  minimum value of width for 50 minima is 50λ

The  minimum value of width for 1000 minima is 1000λ

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