Answer:
1.
109.6 cm , - 1.74 , real
2.
1.5
Explanation:
1.
d₀ = object distance = 63 cm
f = focal length of the lens = 40 cm
d = image distance = ?
using the lens equation
[tex]\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}[/tex]
[tex]\frac{1}{40} = \frac{1}{63} + \frac{1}{d}[/tex]
d = 109.6 cm
magnification is given as
[tex]m = \frac{-d}{d_{o}}[/tex]
[tex]m = \frac{-109.6}{63}[/tex]
m = - 1.74
The image is real
2
d₀ = object distance = a
d = image distance = - (a + 5)
f = focal length of lens = 30 cm
using the lens equation
[tex]\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}[/tex]
[tex]\frac{1}{30} = \frac{1}{a} + \frac{1}{- (a + 5)}[/tex]
a = 10
magnification is given as
[tex]m = \frac{-d}{d_{o}}[/tex]
[tex]m = \frac{- (- (a +5))}{a}[/tex]
[tex]m = \frac{(5 + 10)}{10}[/tex]
m = 1.5
