Respuesta :
Answer:
(a).The location of the final image beyond the converging lens is 67.2 cm.
(b). The magnification is 3.2
Explanation:
Given that,
Object height = 2.2 cm
Object distance = 27 cm
Focal length =16 cm
Distance = 11 cm
Suppose find the location of the final image, beyond the converging lens
We need to calculate the image distance
Using formula of lens
[tex]\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}[/tex]
Put the value into the formula
[tex]\dfrac{1}{-16}=\dfrac{1}{v}+\dfrac{1}{27}[/tex]
[tex]\dfrac{1}{v}=\dfrac{1}{-16}-\dfrac{1}{27}[/tex]
[tex]v=-10.0\ cm,[/tex]
We need to calculate the image distance from converging lens
Using formula of lens
Here, u = -10-11=-21 cm
[tex]\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}[/tex]
Put the value into the formula
[tex]\dfrac{1}{16}=\dfrac{1}{v}+\dfrac{1}{21}[/tex]
[tex]\dfrac{1}{v}=\dfrac{1}{16}-\dfrac{1}{21}[/tex]
[tex]v=67.2\ cm,[/tex]
(b). We need to calculate the magnification of the final image
Using formula of magnification
[tex]m=\dfrac{v}{u}[/tex]
Put the value into the formula
[tex]m=\dfrac{67.2}{-21}[/tex]
[tex]m=-3.2\ cm[/tex]
The magnification is 3.2
Negative sign show the inverted image.
Hence, (a). The location of the final image beyond the converging lens is 67.2 cm.
(b). The magnification is 3.2
