All categories

3 years ago

Camera lenses are described in terms of

Physics

1 answer:

Lesechka [4]3 years ago

6 0

Answer:

The position of the image from the camera lens is approximately 121.1 mm

Explanation:

The given parameters of the lens are;

The specification of the camera lens = 120 mm

Therefore, the focal length of the camera lens, f = 120 mm = 0.12 m

The distance of the object from the camera, $d_o$ = 13 m

The lens equation for finding the position of the image is given as follows;

$\dfrac{1}{f} = \dfrac{1}{d_o} + \dfrac{1}{d_i}$

Where;

$d_i$ = The position of the image from the camera lens

Therefore, by plugging in the known values, we have;

$\dfrac{1}{13} = \dfrac{1}{0.12} + \dfrac{1}{d_i}$

$\dfrac{1}{d_i} = \dfrac{1}{0.12} - \dfrac{1}{13} = \dfrac{13 - 0.12}{0.12 \times 13} = \dfrac{12.88}{1.56} = \dfrac{322}{39}$

$\therefore d_i = \dfrac{39}{322} \approx 0.1211$

The position of the image from the camera lens, $d_i$ = 0.1211 m = 121.1 mm

You might be interested in

Two parallel-plate capacitors have the same plate area. Capacitor 1 has a plate separation twice that of capacitor 2, and the qu

Luba_88 [7]

Answer:

$V_1=8 V_2$

Explanation:

Given that:

Area of the plate of capacitor 1= Area of the plate of capacitor 2=A

separation distance of capacitor 2, $d_2=d$

separation distance of capacitor 1, $d_1=2d$

quantity of charge on capacitor 2, $Q_2=Q$

quantity of charge on capacitor 1, $Q_1=4Q$

We know that the Capacitance of a parallel plate capacitor is directly proportional to the area and inversely proportional to the distance of separation.

Mathematically given as:

$C=\frac{k.\epsilon_0.A}{d}$ .....................................(1)

where:

k = relative permittivity of the dielectric material between the plates= 1 for air

$\epsilon_0 = 8.85\times 10^{-12}\,F.m^{-1}$

From eq. (1)

For capacitor 2:

$C_2=\frac{k.\epsilon_0.A}{d}$

For capacitor 1:

$C_1=\frac{k.\epsilon_0.A}{2d}$

$C_1=\frac{1}{2} [ \frac{k.\epsilon_0.A}{d}]$

We know, potential differences across a capacitor is given by:

$V=\frac{Q}{C}$ ..........................................(2)

where, Q = charge on the capacitor plates.

for capacitor 2:

$V_2=\frac{Q}{\frac{k.\epsilon_0.A}{d}}$

$V_2=\frac{Q.d}{k.\epsilon_0.A}$

& for capacitor 1:

$V_1=\frac{4Q}{\frac{k.\epsilon_0.A}{2d}}$

$V_1=\frac{4Q\times 2d}{k.\epsilon_0.A}$

$V_1=8\times [\frac{Q.d}{k.\epsilon_0.A}]$

$V_1=8 V_2$

6 0

3 years ago

I am trying to find the magnitude of a resultant vector. Do i take inconsideration the negatives when i find the x & y compo

attashe74 [19]

Absolutely ! If you have two vectors with equal magnitudes and opposite
directions, then one of them is the negative of the other. Their correct
vector sum is zero, and that's exactly the magnitude of the resultant vector.

(Think of fifty football players pulling on each end of the rope in a tug-of-war.
Their forces are equal in magnitude but opposite in sign, and the flag that
hangs from the middle of the rope goes nowhere, because the resultant
force on it is zero.)

This gross, messy explanation is completely applicable when you're totaling up
the x-components or the y-components.

4 0

4 years ago

Find the kinetic energy of an electron whose de broglie wavelength is 34.0 nm.

dezoksy [38]

The De Broglie wavelength of the electron is
$\lambda=34.0 nm=34 \cdot 10^{-9} m$
And we can use De Broglie's relationship to find its momentum:
$p= \frac{h}{\lambda}= \frac{6.6 \cdot 10^{-34} Js}{34 \cdot 10^{-9} m}=1.94 \cdot 10^{-26} kg m/s$

Given $p=mv$ , with m being the electron mass and v its velocity, we can find the electron's velocity:
$v= \frac{p}{m}= \frac{1.94 \cdot 10^{-26} kgm/s}{9.1 \cdot 10^{-31} kg}= 2.13 \cdot 10^4 m/s$

This velocity is quite small compared to the speed of light, so the electron is non-relativistic and we can find its kinetic energy by using the non-relativistic formula:
$K= \frac{1}{2}mv^2= \frac{1}{2}(9.1 \cdot 10^{-31} kg)(2.13 \cdot 10^4 m/s)^2=2.06 \cdot 10^{-22} J$

3 0

3 years ago

Two ropes have equal length and are stretched the same way. The speed of a pulse on rope 1 is 1.4 times the speed on rope 2. Par

kondor19780726 [428]

Answer:

m1/m2 = 0.51

Explanation:

First to all, let's gather the data. We know that both rods, have the same length. Now, the expression to use here is the following:

V = √F/u

This is the equation that describes the relation between speed of a pulse and a force exerted on it.

the value of "u" is:

u = m/L

Where m is the mass of the rod, and L the length.

Now, for the rod 1:

V1 = √F/u1 (1)

rod 2:

V2 = √F/u2 (2)

Now, let's express V1 in function of V2, because we know that V1 is 1.4 times the speed of rod 2, so, V1 = 1.4V2. Replacing in the equation (1) we have:

1.4V2 = √F/u1 (3)

Replacing (2) in (3):

1.4(√F/u2) = √F/u1 (4)

Now, let's solve the equation 4:

[1.4(√F/u2)]² = F/u1

1.96(F/u2) =F/u1

1.96F = F*u2/u1

1.96 = u2/u1 (5)

Now, replacing the expression of u into (5) we have the following:

1.96 = m2/L / m1/L

1.96 = m2/m1 (6)

But we need m1/m2 so:

1.96m1 = m2

m1/m2 = 1/1.96

m1/m2 = 0.51

5 0

4 years ago

A person is in a room whose walls are maintained at a temperature of 17 °C. The temperature of the person’s skin is 32 °C. The s

lbvjy [14]

Answer:

sned me short notes of Physics of all branches (post graduation level) thanks

+923466867221

Explanation:

4 0

3 years ago