Question

You have a lens whose focal length is 12.9 cm. You place an object on the axis of the lens at a distance of 6.59 cm from it. How far is the object's image from the lens (as a positive quantity)? Number cm If it can be determined, is the image real or virtual? OCannot be determined Real O Virtual If it can be determined, is the image upright or inverted with respect to the object? O Cannot be determined O Upright O Inverted

Answer 1

Answer:

(I). The distance of the image from the lens is 13.47 cm.

(I). The image is virtual.

(III). The image is upright.

Explanation:

Given that,

Focal length = 12.9 cm

Object distance = 6.59 cm

(I). We need to calculate the image distance

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

Where, f = focal length

v = image distance

u = object distance

Put the value into the formula

$\dfrac{1}{12.9}=\dfrac{1}{v}-\dfrac{1}{-6.59}$

$\dfrac{1}{v}=\dfrac{1}{12.9}-\dfrac{1}{6.59}$

$\dfrac{1}{v}=-\dfrac{6310}{85011}$

$v=-13.47\ cm$

The distance of the image from the lens is 13.47 cm.

(II). Negative sign shows the image is formed in left side of the lens.

So, the image is virtual.

(III). We need to calculate the magnification

Using formula of magnification

$m=\dfrac{v}{u}$

Put the value into the formula

$m=\dfrac{-13.47}{-6.59}$

$m =2.04$

Positive sign show the image is upright.

Hence, (I). The distance of the image from the lens is 13.47 cm.

(II). The image is virtual.

(III). The image is upright.