Question 928c3

Nov 21, 2017

It is a real, inverted image that is 1.5 times bigger than the object and forms on the side opposite to that of the object.

Explanation:

Thin Lens Equation: $\setminus q \quad \frac{1}{u} + \frac{1}{v} = \frac{1}{f}$ ...... (1)
$u$ : is the object distance (always positive)
$v$ : is the image distance (positive if the image forms on the opposite side as the object)
$f$ : is the focal length of the lens (positive for a converging lens)

Given: $\setminus q \quad u = + 60$ cm; \qquad f = +20# $c m$

Image Distance: Invert (1) to find the image distance $v$

$v = {\left(\frac{1}{f} - \frac{1}{u}\right)}^{- 1} = {\left(\frac{1}{20 c m} - \frac{1}{60 c m}\right)}^{- 1} = + 30$ $c m$

Positive image distance indicates that the image is a real image and it forms on the side that is opposite to that of the object.

Magnification: $\setminus q \quad m = - \frac{v}{u} = - \frac{30 c m}{20 c m} = - 1.5$

Negative sign for magnification indicates that the image is inverted. Its magnitude shows that the image is 1.5 times the size of the object.