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 = ![\frac{1}{f}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%7D)
where
f = focal length
Thus
f = ![\frac{1}{P}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BP%7D)
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:
![\frac{1}{f} = \frac{1}{u} + \frac{1}{u'}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%7D%20%3D%20%5Cfrac%7B1%7D%7Bu%7D%20%2B%20%5Cfrac%7B1%7D%7Bu%27%7D)
where
u = object distance = 25 cm = 0.25 m = near point of a normal eye
u' = image distance
Now,
![\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bu%27%7D%20%3D%20%5Cfrac%7B1%7D%7Bf%7D%20-%20%5Cfrac%7B1%7D%7Bu%7D)
![\frac{1}{u'} = \frac{1}{0.5} - \frac{1}{0.25}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bu%27%7D%20%3D%20%5Cfrac%7B1%7D%7B0.5%7D%20-%20%5Cfrac%7B1%7D%7B0.25%7D)
![\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bu%27%7D%20%3D%20%5Cfrac%7B1%7D%7Bf%7D%20-%20%5Cfrac%7B1%7D%7Bu%7D)
Solving the above eqn, we get:
u' = - 0.5 m = - 50 cm
Answer:
The mass of the man is 71 kg
Explanation:
Given;
kinetic energy of the man, K.E = 887.5 J
velocity of the man, v = 5 m/s
The mass of the man is calculated as follows;
K.E = ¹/₂mv²
where;
m is the mass of the man
2K.E = mv²
m = 2K.E / v²
m = (2 x 887.5) / (5)²
m = 71 kg
Therefore, the mass of the man is 71 kg
Answer:
![5.04\cdot 10^8 A](https://tex.z-dn.net/?f=5.04%5Ccdot%2010%5E8%20A)
Explanation:
The work function of the metal corresponds to the minimum energy needed to extract a photoelectron from the metal. In this case, it is:
![\phi = 3.950\cdot 10^{-19}J](https://tex.z-dn.net/?f=%5Cphi%20%3D%203.950%5Ccdot%2010%5E%7B-19%7DJ)
So, the energy of the incoming photon hitting on the metal must be at least equal to this value.
The energy of a photon is given by
![E=\frac{hc}{\lambda}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7Bhc%7D%7B%5Clambda%7D)
where
h is the Planck's constant
c is the speed of light
is the wavelength of the photon
Using
and solving for
, we find the maximum wavelength of the radiation that will eject electrons from the metal:
![\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{3.950\cdot 10^{-19} J}=5.04\cdot 10^{-7}m](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7Bhc%7D%7BE%7D%3D%5Cfrac%7B%286.63%5Ccdot%2010%5E%7B-34%7D%20Js%29%283%5Ccdot%2010%5E8%20m%2Fs%29%7D%7B3.950%5Ccdot%2010%5E%7B-19%7D%20J%7D%3D5.04%5Ccdot%2010%5E%7B-7%7Dm)
And since
1 angstrom = ![10^{-15}m](https://tex.z-dn.net/?f=10%5E%7B-15%7Dm)
The wavelength in angstroms is
![\lambda=\frac{5.04\cdot 10^{-7} m}{10^{-15} m/A}=5.04\cdot 10^8 A](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7B5.04%5Ccdot%2010%5E%7B-7%7D%20m%7D%7B10%5E%7B-15%7D%20m%2FA%7D%3D5.04%5Ccdot%2010%5E8%20A)
The equation for the de Broglie wavelength is:
<span>λ = (h/mv) √[1-(v²/c²)], </span>
<span>where h is Plank's Constant, m is the rest mass, v is velocity, and c is the velocity of light in vacuum. However, if c>>v (and it is, in this case) then the expression under the radical sign approaches 1, and the equation simplifies to: </span>
<span>λ = h/mv. </span>
<span>Substituting, (remember to convert the mass to kg, since 1 J = 1 kg·m²/s²): </span>
<span>λ = (6.63x10^-34 J·s) / (0.0459 kg) (72.0 m/s) = 2.00x10^-34 m.</span>