Answer:
a) Focal length of the lens is 8 cm which is a convex lens
b) 6 cm
c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.
Explanation:
u = Object distance = 4 cm
v = Image distance = -8 cm
f = Focal length
Lens Equation
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}=\frac{1}{4}+\frac{1}{-8}\\\Rightarrow \frac{1}{f}=\frac{1}{8}\\\Rightarrow f=\frac{8}{1}=-8\ cm[/tex]
a) Focal length of the lens is 8 cm which is a convex lens
Magnification
[tex]m=-\frac{v}{u}\\\Rightarrow m=-\frac{-8}{4}\\\Rightarrow m=2[/tex]
b) Height of image is 2×3 = 6 cm
Since magnification is positive the image upright
c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.
(a) The focal length of the lens is -2.67 cm.
(b) The magnification of the image is 2 and the height is 6 cm.
(c) The imaged formed by the lens is upright, virtual and magnified.
The focal length of the lens is determined by using lens formulas as given below;
[tex]\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\\\\frac{1}{f} = \frac{-1}{8} - \frac{1}{4} \\\\\frac{1}{f} = \frac{-1 -2}{8} = \frac{-3}{8} \\\\f = -2.67 \ cm[/tex]
The magnification of the image is calculated as follows;
[tex]m = \frac{v}{u} \\\\m = \frac{8}{4} \\\\m = 2\\\\[/tex]
The height of the image is calculated as follows;
[tex]H = mu\\\\H = 2(3cm) = 6\ cm[/tex]
The imaged formed by the lens is;
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