Physics, 14.09.2019 07:10 dondre54

Ordinary glasses are worn in front of the eye and usually 2.00 cm in front of the eyeball. a certain person can see distant objects well, but his near point is 50.0 cm from his eyes instead of the usual 25.0 cm . suppose that this person needs ordinary glasses what focal length lenses are needed to correct his vision ? what is their power in diopters?

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Answer from: thompsonjodi9638

The focal length is 15.549 cm

The power of the lens is 0.0643 D

Solution:

As per the question:

The near point is 50.0 cm

Distance of the glasses from the eyeball, d = 2.00 cm

The near point of a normal human eye is 25 cm

Now,

The image distance, v' = 50.0 - 2.00 = 48.0 cm

The object distance, u' = 25.0 - 2.00 = 23.0 cm

Now, using the Lens maker formula to calculate the focal length:

$\frac{1}{f} = \frac{1}{u'} + \frac{1}{v'}$

$\frac{1}{f} = \frac{1}{48.0} + \frac{1}{23.0} = 0.0643$

f = 15.549 cm

Now, the power of the lens in diopters is given by:

$P = \frac{1}{f} = \frac{1}{15.549} = 0.0643 D$

Answer from: Quest

answer: the correct answer is 1.

explanation: there are various process in which a radioactive nuclei decays:

1) alpha decay: in this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. the mass number is reduced by 4 units.

$_a^z\textrm{x}\rightarrow _{a-2}^{z-4}+_2^4\alpha$

2) electron capture: in this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. simply, a proton and an electron combines together to form a neutron. mass number does not change in this process.

$_a^z\textrm{x}+_{-1}^0\textrm{e}\rightarrow _{z-1}^a\textrm{y}$

3) gamma ray emission: it is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. this decay releases $\gamma -radiations$ . this process does not change the mass number.

$_a^z\textrm{x}^*\rightarrow _a^z\textrm{x}+_0^0\gamma$

4) positron emission: it is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. this is also known as $\beta ^+$ -decay. in this the mass number remains same.

$_a^z\textrm{x}\rightarrow _{z-1}^a\textrm{y}+_{+1}^0e$

from the above information, we see that only alpha-decay is producing a change in the mass number of the nucleus.

Answer from: Quest

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic = - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day

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