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2 years ago

the person is not able to see the far lying object clearly. which type of lens will the doctor prescribe him give reason?

Physics

1 answer:

ASHA 777 [7]2 years ago

8 0

Answer:

correction of farsightedness uses a converging lenses that compensates for the under convergence by the eye. the converging lens produces an image farther from the eye than the object so that farsighted can see it clearly.

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How many hydrogen and carbon atoms in a diamond

Wewaii [24]

Answer:

Explanation:

Thus, total 4+4=8 C atoms are present per unit cell of diamond. Carbon has an electronic arrangement of 2,4. In diamond, each carbon shares electrons with four other carbon atoms - forming four single bonds.

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2 years ago

A monochromatic light beam is incident on a barium target that has a work function of 2.50 \mathrm{eV} . If a potential differen

leva [86]

The wavelength of the light beam required to turn back all the ejected electrons is 497 nm which is option (b).

Work function is a material property defined as the minimum amount of energy required to infinitely remove electrons from the surface of a particular solid.
The potential difference required to support all emitted electrons is called the stopping potential which is given by $v_0=\frac{K.E_m_a_x}{e}$ .....(1)
where $v_0$ is the stopping potential and e is the charge of the electron given by $1.6\times10^-^1^9$ .

It is given that work function (Ф) of monochromatic light is 2.50 eV.

Einstein photoelectric equation is given by:

$K.E_m_a_x=E-\phi$ ....(2)

where K.E(max) is the maximum kinetic energy.

Substituting (1) into (2) , we get

$ev_0=E-\phi\\1.6\times10^{-19} \times1=E-2.50\\E=1.6\times10^{-19}+2.50\\E=2.50eV$

As we know that $E=\frac{hc}{\lambda}$ ....(3)

where Speed of light, $c = 3\times10^8 m/s$ and Planck's constant , $h = 6.63\times 10^-^1^9Js = 4.14\times 10^-^1^5 eVs$

From equation (3) , we get

$\lambda=\frac{hc}{E} \\\\\lambda=\frac{ 4.14\times 10^-^1^5 \times 3 \times10^8}{2.50} \\\\\lambda=\frac{1240\times10^-^9}{2.50} \\\\\lambda=496.8\times10^-^9\\\\\lambda=497nm$

Learn about more einstein photoelectric equation here:

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8 months ago

An electric dipole consisting of charges of magnitude 1.70 nC separated by 6.80 μm is in an electric field of strength 1160 N/C.

bazaltina [42]

Answer:

p = 1.16 10⁻¹⁴ C m and ΔU = 2.7 10 -11 J

Explanation:

The dipole moment of a dipole is the product of charges by distance

p = 2 a q

With 2a the distance between the charges and the magnitude of the charges

p = 1.7 10⁻⁹ 6.8 10⁻⁶

p = 1.16 10⁻¹⁴ C m

The potential energie dipole is described by the expression

U = - p E cos θ

Where θ is the angle between the dipole and the electric field, the zero value of the potential energy is located for when the dipole is perpendicular to the electric field line

Orientation parallel to the field

θ = 0º

U = 1.16 10⁻¹⁴ 1160 cos 0

U1 = 1.35 10⁻¹¹ J

Antiparallel orientation

θ = 180º

cos 180 = -1

U2 = -1.35 10⁻¹¹ J

The difference in energy between these two configurations is the subtraction of the energies

ΔU = | U1 -U2 |

ΔU = 1.35 10-11 - (-1.35 10-11)

ΔU = 2.7 10 -11 J