What is the object distance? you will need to use the magnification equation to find a relationship between s and s′. then substitute into the thin lens equation to solve for s.?

Respuesta :

 For lens in S I system,1/v - 1 /u = 1/f 

gives 1/v=1/f + 1/u, 

multiply by 'u' on both sides 

u/v =u/f + 1 

u/v = ( u+f ) / f 

inverting, 

v/u = f / (u+f) 

Therefore magnification 'm' = v/u=f / (u+f) 

'm' = f / (u+f) 

Given magnification is 5 / 9 

f / (u+f) = 5 / 9 

object distance 'u'= 4f /5 

Now substitute for focal length f ( focal length of diverging lens is negative)

The distance of the object, when the value of s' is substitute into the thin lens equation to solve for s is 12 cm.

What is focal length of the lens?

The focal length of the lens is length of the distance between the middle of the lens to the focal point.

It can be find out using the following formula as,

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

Here, (v)is the distance of the image, (u) is the distance of the object, and (f) is the focal length of the lens.

Using the magnification equation, the relationship between s and s′ can be given as,

[tex]m=\dfrac{s'}{s}[/tex]

Here, s' is the image distance and s is the object distance. The value of magnification is 5/4, obtained in previous part and the image distance is 15 cm given in the problem. Thus,

[tex]\dfrac{5}{4}=\dfrac{15}{s}\\s=12\rm\; cm[/tex]

Thus, the distance of the object, when the value of s' is substitute into the thin lens equation to solve for s is 12 cm.

Learn more about the focal length here;

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