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The bellows of an adjustable camera can be extended so that the largest distance from the lens to the film is 1.45 times the focal length. If the focal length of the lens is 6.14 cm, what is the distance from the closest object that can be sharply focused on the film

Sagot :

Answer: 19.80 cm

Explanation:

Given

focal length [tex]f=6.14\ cm[/tex]   (as focal length is positive, it is converging lens)

Image distance [tex]v=1.5f[/tex]

[tex]v=1.5\times 6.14\\v=9.21\ cm[/tex]

using lens formula

[tex]\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}[/tex]

insert values

[tex]\dfrac{1}{u}=\dfrac{1}{6.14}-\dfrac{1}{8.9}\\\\\dfrac{1}{u}=0.1628-0.1123\\\\\dfrac{1}{u}=0.0505\\\\u=19.80\ cm[/tex]

Thus, the distance of the object is 19.80 cm

Cricetus

The distance of the object will be "19.80 cm".

Given:

  • Focal length, f = 6.14 cm

Now,

The image distance will be:

→ [tex]v = 1.5 f[/tex]

     [tex]= 1.5\times 6.14[/tex]

     [tex]= 9.21 \ cm[/tex]

By using the lens formula, we get

→ [tex]\frac{1}{v} + \frac{1}{u} = \frac{1}{f}[/tex]

By putting the values, we get

→ [tex]\frac{1}{u} = \frac{1}{6.14} - \frac{1}{8.9}[/tex]

   [tex]\frac{1}{u} = 0.1628-0.1123[/tex]

   [tex]\frac{1}{u} = 0.0505[/tex]

   [tex]u = 19.80 \ cm[/tex]

Thus the answer above is correct.    

Learn more about focal length here:

https://brainly.com/question/25020197

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