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Tatiana [17]
4 years ago
9

An object is placed 18 cm in front of spherical mirror.if the image is formed at 4cm to the right of the mirror, calculate it's

focal length. Is the mirror convex or concave? What is the nature of the image? What is the radius of curvature of the mirror
Physics
1 answer:
ivolga24 [154]4 years ago
3 0
1) Focal length

We can find the focal length of the mirror by using the mirror equation:
\frac{1}{f}= \frac{1}{d_o}+ \frac{1}{d_i} (1)
where 
f is the focal length
d_o is the distance of the object from the mirror
d_i is the distance of the image from the mirror

In this case, d_o = 18 cm, while d_i=-4 cm (the distance of the image should be taken as negative, because the image is to the right (behind) of the mirror, so it is virtual). If we use these data inside (1), we find the focal length of the mirror:
\frac{1}{f}= \frac{1}{18 cm}- \frac{1}{4 cm}=- \frac{7}{36 cm}
from which we find
f=- \frac{36}{7} cm=-5.1 cm

2) The mirror is convex: in fact, for the sign convention, a concave mirror has positive focal length while a convex mirror has negative focal length. In this case, the focal length is negative, so the mirror is convex.

3) The image is virtual, because it is behind the mirror and in fact we have taken its distance from the mirror as negative.

4) The radius of curvature of a mirror is twice its focal length, so for the mirror in our problem the radius of curvature is:
r=2f=2 \cdot 5.1 cm=10.2 cm
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Answer:

a)n =7 , b) n = 5

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This equation energy is given in elector volts and n is an integer

A transition occurs when the electro sees from a superior to a lower state

       E₀ - E_{n} = -13.606 (1/n_{f}² - 1/n₀²)

Let's apply this expression

         n₀ = 2

Let's look for the energy of the different levels and subtract it

n₀          E_{n} (eV)

1            -13,606

2            -3.4015

3           - 1.5118

4           -0.850375

5           -0.54424

6           -0.3779

7           -0.2777

The wavelength of the transition is 397 nm = 397 10⁻⁹ m

The speed of light is related to wavelength and frequency

       c = λ f

The Planck equation gives the energy of a transition

      E = h f

      E = h c /λ

Let's calculate

     E = 6.63 10⁻³⁴ 3 10⁸/397 10⁻⁹

     E = 5.01 10⁻¹⁹ J

Let's reduce to eV

     E = 5.01 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)

     E = 3.1313 eV

Let's examine the possible transitions from the initial level ni = 2

     ΔE = E_{2} - E_{n} = -3.1313

     En = 3.13 -3.4015

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When examining the table we see that the level that this energy has is the level of n = 7

Part B      the transition is in the infrared

The frequency is 74 10¹² Hz

We use the Planck equation

       E = h f

       E = 6.63 10⁻³⁴ 74 10¹²

       E = 4.9062 10⁻²⁰ J

       E = 4.9062 10⁻²⁰ / 1.6 10⁻¹⁹

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We look for the level with the energy difference

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