All categories

2 years ago

The power of a lens is 4. 0 diopters and its diameter is 5. 0 cm. what is the focal length of this lens?

Physics

1 answer:

fenix001 [56]2 years ago

8 0

The focal length of the lens is 25cm

given:

power,p=4 diopters

what is focal length?

Focal length is the distance between the point of convergence of your lens and the sensor or film recording the image.

what is diopter?

The unit of optical power of lens is called diopter.It is the optical power of the lens.

we know,

p=1/f

where,

p= power

f= focal length

f=1/p

f=1/4

=0.25m

=25cm

Thus,the focal length of the lens is 25cm

learn more about focal length from here: https/brainly.com/question/28203589

#SPJ4

You might be interested in

Based on the graph, which data point is most likely to have experimental

Len [333]

Answer:

B. 59 kg

Explanation:

From the graph you notice that a linear relation in indicated by the line joining the points such that the points on the line represent the data that show a correct relationship in the experiment.

This means that the point outside the line has an error .

This point is the value 59 kg that does not align with other values which are included in the graph.

8 0

3 years ago

Using only one management style with all people is the most effective leadership technique for any organization.

Softa [21]

the answer for this is false

4 0

3 years ago

One of the main factors driving improvements in the cost and complexity of integrated circuits (ICs) is improvements in photolit

nika2105 [10]

Answer:

0.000003782 m

0.000001891 m

0.000001197125 m

Explanation:

$\lambda$ = Wavelength = 248 nm

D = Diameter of beam = 1 cm

f = Focal length = 0.625 cm

The angle is given by

$\theta=\dfrac{1.22\lambda}{D}$

The width is given by

$d=2\theta f\\\Rightarrow d=2\dfrac{1.22\lambda f}{D}\\\Rightarrow d=2\dfrac{1.22\times 248\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.000003782\ m$

The required width is 0.000003782 m

Minimum resolvable line separation is given by

$\dfrac{0.000003782}{2}=0.000001891\ m$

The minimum resolvable line separation between adjacent lines is 0.000001891 m

when $\lambda=157\ nm$

$d=2\dfrac{1.22\times 157\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.00000239425\ m$

The new minimum resolvable line separation between adjacent lines is

$\dfrac{0.00000239425}{2}=0.000001197125\ m$

6 0

4 years ago

A figure skater glides along a circular path of radius 6.70 m. If she coasts around one half of the circle, find the following.

adoni [48]

(a) Her distance from the starting location is 21.05 m.

(b) The length of the path she skated is 21.05 m.

<h3>Distance of the skater from the starting position</h3>

The distance around a complete circular path is calculated as 2πr.

The distance for a half circle is calculated as ¹/₂ x 2πr = πr

Distance from the starting location = π x 6.7 m = 21.05 m

The length of the path she skated is the same as her distance from the starting location = 21.05 m.

Learn more about distance round a circle here: brainly.com/question/3100527

#SPJ1

8 0

2 years ago

Vector vector b has x, y, and z components of 4.00, 4.00, and 2.00 units, respectively. calculate the magnitude of vector

Sav [38]

Good morning.

We see that $\mathsf{\overset{\to}{b}} = \mathsf{(4.00, \ 4.00, \ 2.00)}$

The magnitude(norm, to be precise) can be calculated the following way:

$\star \ \boxed{\mathsf{\overset{\to}{a}=(x, y,z)\Rightarrow ||\overset{\to}{a}|| = \sqrt{x^2+y^2+z^2}}}$

Now the calculus is trivial:

$\mathsf{\|\overset{\to}{b} \| =\sqrt{4^2+4^2+2^2} =\sqrt{16+16+4}}\\ \\ \mathsf{\|\overset{\to}{b}\|=\sqrt{36}}\\ \\ \boxed{\mathsf{\|\overset{\to}{b}\| = 6.00 \ u}}$

7 0

3 years ago