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An object of 8 cm height is 15 cm in front of a converging lens with focal length of magnitude 4 cm. What is the dimension of the image? Do not include the sign)?

Respuesta :

Answer:

image height is 2.9 cm

Explanation:

We have given the size of the object O = 8 cm

Distance of the object from the lens = 15 cm

Focal length of the lens = 4 cm

Now for lens we know that [tex]\frac{1}{f}=\frac{1}{v}-\frac{1}{u}[/tex]

So [tex]\frac{1}{4}=\frac{1}{v}-\frac{1}{15}[/tex]

v = 5.454 cm

Now magnification [tex]m=\frac{v}{u}=\frac{5.454}{15}=0.3636[/tex]

Now magnification is the ratio of image height and object height

So [tex]m=\frac{I}{O}[/tex] , where I is the height of image and O is the height of object

So [tex]0.3636=\frac{I}{8}[/tex]

I =2.9 cm

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