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Answer:
The near point of an eye with power of +2 dopters, u' = - 50 cm
Given:
Power of a contact lens, P = +2.0 diopters
Solution:
To calculate the near point, we need to find the focal length of the lens which is given by:
Power, P =
where
f = focal length
Thus
f =
f = = + 0.5 m
The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.
Now, by using lens maker formula:
where
u = object distance = 25 cm = 0.25 m = near point of a normal eye
u' = image distance
Now,
Solving the above eqn, we get:
u' = - 0.5 m = - 50 cm
To develop this problem it is necessary to apply the concepts related to Wavelength, The relationship between speed, voltage and linear density as well as frequency. By definition the speed as a function of the tension and the linear density is given by
Where,
T = Tension
Linear density
Our data are given by
Tension , T = 70 N
Linear density ,
Amplitude , A = 7 cm = 0.07 m
Period , t = 0.35 s
Replacing our values,
Speed can also be expressed as
Re-arrange to find \lambda
Where,
f = Frequency,
Which is also described in function of the Period as,
Therefore replacing to find
Therefore the wavelength of the waves created in the string is 3.49m