A 3.0-cm tall statue is 48 cm in front of a convex lens having a focal length of i. (2 pts) Is the image of the statue real or virtual (give reason (2 pts) for your answer)? magnitude 20 cm. ii. (5 pts) How far is the image from the lens? ii. 3 pts) ls the image upright or inverted? iv. (5 pts) How tall is the image?

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CarliReifsteck CarliReifsteck · Answer 1 · 2019-09-04T14:14:43+02:00

Explanation:

Given that,

Height of object = 3.0 cm

Distance of object u= 48 cm

Focal length = 48 cm

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}{20}=\dfrac{1}{v}+\dfrac{1}{-48}[/tex]

[tex]\dfrac{1}{v}=\dfrac{1}{20}+\dfrac{1}{48}[/tex]

[tex]\dfrac{1}{v}=\dfrac{17}{240}[/tex]

[tex]v=14.11\ cm[/tex]

(I). The image is real.

(II). The distance of the image from the lens is 14.11 cm.

(II). The image is inverted.

(IV). We need to calculate the height of the image

Using formula of magnification

[tex]m = \dfrac{v}{u}=\dfrac{h'}{h}[/tex]

[tex]\dfrac{v}{u}=\dfrac{h'}{h}[/tex]

Put the value into the formula

[tex]\dfrac{14.11}{48}=dfrac{h'}{3.0}[/tex]

[tex]h'=\dfrac{14.11}{48}\times3.0[/tex]

[tex]h'=0.88\ cm[/tex]

The height of the image is 0.88 cm.

Hence, This is the required solution.