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3 years ago

How does the thickness of the lens affects its focal length plz ans this i will mark as brainliest

Physics

1 answer:

Oksanka [162]3 years ago

5 0

Hello! :)

The focal length of the lens tells you how far away from the lens a focused image is created, if light rays approaching the lens are parallel. A lens with more “bending power” has a shorter focal length, because it alters the path of the light rays more effectively than a weaker lens. Most of the time, you can treat a lens as being thin and ignore any effects from the thickness, because the thickness of the lens is much less than the focal length. But for thicker lenses, how thick they are does make a difference, and in general, results in a shorter focal length.

Hope I helped and didn’t answer too late!

Good luck and stay COOL!

~ Destiny ^_^

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A mass of 0.40 kg is suspended on a spring which then stretches 10 cm. The mass is then removed and a second mass is placed on t

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Answer:

x' = 1.01 m

Explanation:

given,

mass suspended on the spring, m = 0.40 Kg

stretches to distance, x = 10 cm = 0. 1 m

now,

we know

m g = k x

where k is spring constant

0.4 x 9.8 = k x 0.1

k = 39.2 N/m

now, when second mass is attached to the spring work is equal to 20 J

work done by the spring is equal to

$W = \dfrac{1}{2}kx'^2$

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7 0

3 years ago

A laser emits two wavelengths (λ1 = 420 nm; λ2 = 630 nm). When these two wavelengths strike a grating with 450 lines/mm, they pr

Westkost [7]

A) Order of the first laser: 3, order of the second laser: 2

B) The overlap occurs at an angle of $34.9^{\circ}$

Explanation:

The formula that gives the position of the maxima (bright fringes) for a diffraction grating is

$d sin \theta = m \lambda$

where

d is spacing between the lines in the grating

$\theta$ is the angle of the maximum

m is the order of diffraction

$\lambda$ is the wavelength of the light

For laser 1,

$d sin \theta = m_1 \lambda_1$

For laser 2,

$d sin \theta = m_2 \lambda_2$

where

$\lambda_1 = 420 nm\\\lambda_2 = 630 nm$

Since the position of the maxima in the two cases overlaps, then the term $d sin \theta$ on the left is the same for the two cases, therefore we can write:

$m_1 \lambda_1 = m_2 \lambda_2\\\frac{m_1}{m_2}=\frac{\lambda_2}{\lambda_1}=\frac{630}{420}=\frac{3}{2}$

Therefore:

$m_1 = 3$

$m_2 = 2$

In order to find the angle at which the overlap occurs, we use the 1st laser situation:

$d sin \theta = m_1 \lambda_1$

where:

N = 450 lines/mm = 450,000 lines/m is the number of lines per unit length, so the spacing between the lines is

$d=\frac{1}{N}=\frac{1}{450,000}=2.2\cdot 10^{-6} m$

$m_1 = 3$ is the order of the maximum

$\lambda_1 = 420 nm = 420\cdot 10^{-9} m$ is the wavelength of the laser light

Solving for $\theta$ , we find the angle of the maximum:

$sin \theta = \frac{m_1 \lambda_1}{d}=\frac{(3)(420\cdot 10^{-9})}{2.2\cdot 10^{-6}}=0.572$

So the angle is

$\theta=sin^{-1}(0.572)=34.9^{\circ}$

Learn more about diffraction:

brainly.com/question/3183125

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3 years ago

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Answer:

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3 years ago

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sattari [20]

Answer: a

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3 years ago

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kondor19780726 [428]

Answer:

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