Answer:
f = 19,877 cm and P = 5D
Explanation:
This is a lens focal length exercise, which must be solved with the optical constructor equation
1 / f = 1 / p + 1 / q
where f is the focal length, p is the distance to the object and q is the distance to the image.
In this case the object is placed p = 25 cm from the eye, to be able to see it clearly the image must be at q = 97 cm from the eye
let's calculate
1 / f = 1/97 + 1/25
1 / f = 0.05
f = 19,877 cm
the power of a lens is defined by the inverse of the focal length in meters
P = 1 / f
P = 1 / 19,877 10-2
P = 5D
Answer:
0.22mm
Explanation:
A far sighted person is a person suffering from long sightedness i.e such individual can only see far distant object clearly but not near distant object. The defect is corrected using convex lens.
Since convex lens is used, the focal (f) length of the lens is positive and the image distance (v) is also positive.
Using the lens formula,
1/f = 1/u + 1/v
Where u is the object distance = 0.35mm
v = 0.6mm
1/f = 1/0.35+1/0.6
1/f = 2.86 + 1.67
1/f = 4.53
f = 1/4.53
f = 0.22mm
The focal length of the contact lenses will be 0.22mm
Answer:
180 Newton(N)
Explanation:
force =mass *acceleration
=60 * 3
=180 kgm/s^2
=180 N
Explanation:
The time taken by a wave crest to travel a distance equal to the length of wave is known as wave period.
The relation between wave period and frequency is as follows.
T = \frac{1}{f}T=
f
1
where, T = time period
f = frequency
It is given that wave period is 18 seconds. Therefore, calculate the wave period as follows.
T = \frac{1}{f}T=
f
1
or, f = \frac{1}{T}f=
T
1
= \frac{1}{18 sec}
18sec
1
= 0.055 per second (1cycle per second = 1 Hertz)
or, f = 5.5 \times 10^{-2} hertz5.5×10 −2 hertz
<h3>Thus, we can conclude that the frequency of the wave is 5.5 \times 10^{-2} hertz5.5×10 −2 hertz .</h3>
