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Pani-rosa [81]
2 years ago
10

The focal length of a diverging lens is negative. If f = −26 cm or a particular diverging lens, where will the image be formed o

f an object located 63 cm to the left of the lens on the optical axis?
.... cm to the left of the lens

....What is the magnification of the image?
Physics
1 answer:
Arada [10]2 years ago
7 0

The distance of the image formed is 44.27 cm and the magnification of the image is 0.7.

<h3>Location of the image</h3>

The image distance formed by the lens is calculated as follows;

1/v + 1/u = -1/f

where;

  • v is the image distance
  • u is the object distance
  • f is the focal length of the lens

1/v = -1/f - 1/u

1/v = -(-1/26) - 1/63

1/v = 1/26 - 1/63

1/v = 0.022588

v = 1/0.022588

v = 44.27 cm

<h3>What is magnification of lens?</h3>

The magnification of a lens is defined as the ratio of the height of an image to the height of an object.

<h3>Magnification of the image formed</h3>

The magnification of the image is calculated as follows;

Magnification = image distance/object distance

M = 44.27/63

M = 0.7

Thus, the distance of the image formed is 44.27 cm and the magnification of the image is 0.7.

Learn more about magnification here: brainly.com/question/1599771

#SPJ1

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The given parameters;

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The vertical component of the initial velocity is given as;

y = v_0_yt - \frac{1}{2} gt^2\\\\v_0_yt = y + \frac{1}{2} gt^2\\\\v_0_y = \frac{y}{t} + \frac{1}{2} gt

The horizontal component of the initial velocity is calculated as;

x = v_0_xt\\\\v_0_x = \frac{x}{t}

The time to reach to the maximum height is calculated as;

T = \frac{v_f_y -v_0_y}{-g} \\\\T = \frac{-v_0_y}{-g} \\\\T = \frac{v_0_y}{g} \\\\T =  \frac{1}{g}  (v_0_y)\\\\T = \frac{1}{g} (\frac{y}{t} + \frac{1}{2} gt)\\\\T = \frac{y}{gt} + \frac{1}{2} t

The horizontal displacement when the object reaches maximum height is calculated as;

X= v_0_x \times T\\\\X= \frac{x}{t} \times (\frac{y}{gt} + \frac{1}{2} t)\\\\X = \frac{xy}{gt^2} + \frac{x}{2}

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