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MaRussiya [10]
3 years ago
9

Which of the following diagrams involves a virtual image ?

Physics
1 answer:
sergiy2304 [10]3 years ago
7 0

Answer:

The third diagram

Explanation:

  • <u>A virtual image</u> is an image that can not be formed on a screen.
  • <u>A convex lens</u> can form both virtual and real image depending on the position of the object from the lens.
  • A virtual image in convex lens is formed when the object is placed between the focus and the optical center of the lens.
  • In the third diagram, a virtual image is formed because the position of the object is between the focus and the optical center of the convex lens.
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What is the strength of electric field EpEp 0.60 mmmm from a proton? Express your answer to two significant figures and include
disa [49]

Answer:

3.99*10^-3N/C

Explanation:

Using

Ep= kq/r²

Where r = 0.6mm = 0.6*10^-3m

K= 8.9*10^9 and q= 1.6*10^-19

So = 8.9*10^9 * 1.6*10^-19/0.6*10^-3)²

= 3.99*10^-3N/C

8 0
3 years ago
In a double-slit interference experiment, the slit separation is 2.41 μm, the light wavelength is 512 nm, and the separation bet
Elina [12.6K]

Answer:

Using equation 2dsinФ=n*λ

given d=2.41*10^-6m

λ=512*10^-12m

θ=52.64 degrees

7 0
4 years ago
Read 2 more answers
Reading glasses of what power are needed for a person whose near point is 125 cm , so that he can read a computer screen at 54 c
Burka [1]

Answer:

1.11 dioptre

Explanation:

d_{i} = Distance of the image = - (125 - 2) = - 123 cm

d_{o} = Distance of the object = 54 - 2 = 52 cm

f = Focal length of the lens

Using the equation

\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{1}{f}

\frac{1}{52} + \frac{- 1}{123} = \frac{1}{f}

f = 90.1 cm

Power of the lens is given as

P = \frac{100}{f}

P = \frac{100}{90.1}

P = 1.11 Dioptre

3 0
3 years ago
How fast does light year travel
KIM [24]
"Light year" is a distance, not a speed. It's the distance light travels in one year, at the speed of 299,792,458 meters per second.
5 0
3 years ago
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A wildlife photographer uses a moderate telephoto lens of focal length 135 mm and maximum aperture f/4.00 to photograph a bear t
labwork [276]

Answer:

<h3>(A) The width x = 0.24 \times 10^{-3} m</h3><h3>(B) The new width is 1.32 \times 10^{-3} m</h3>

Explanation:

Given :

Focal length f =   135 \times 10^{-3}  m

Maximum aperture D = \frac{f}{4}

Wavelength \lambda = 550 \times 10^{-9} m

(A)

From rayleigh criterion,

  \theta = \frac{1.22 \lambda }{D}

  \theta =\frac{ 1.22 \times 550 \times 10^{-9}  }{33.75 \times 10^{-3} }

  \theta = 1.98 \times 10^{-5} rad

From angle formula,

  x = R\theta

Where R = 12 m ( given in example )

x = 12 \times 1.98 \times 10^{-5} m

x = 23.76 \times 10^{-5}

x = 0.24 \times 10^{-3} m

(B)

We know that \theta is proportional to the x and inversely proportional to the D

so we write the new width, here x is 5.5 times larger than above case

   x = 0.24 \times  10^{-3}  \times \frac{22}{4}

   x = 1.32 \times 10^{-3} m

6 0
3 years ago
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