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We have that the magnification of each focal length is given respectively as
A) has [tex]u=3\frac{f}{2}[/tex]
B) has [tex]u=4\frac{f}{3}[/tex]
C) has [tex]u=5\frac{f}{4}[/tex]
From the question we are told that:
Focal Length F
Generally, the equation for Magnification is mathematically given by
[tex]M=\frac{-v}{u}[/tex]
Therefore
[tex]v=2u[/tex]
For A
[tex]M=-2[/tex]
Therefore
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}[/tex]
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{2u}[/tex]
Therefore
[tex]u=3\frac{f}{2}[/tex]
For B
[tex]M=-3[/tex]
Therefore
[tex]v=3u[/tex]
Where
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}[/tex]
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{3u}[/tex]
Therefore
[tex]u=4\frac{f}{3}[/tex]
For C
[tex]M=-4[/tex]
Therefore
[tex]v=4u[/tex]
Therefore
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}[/tex]
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{4u}[/tex]
Therefore
[tex]u=5\frac{f}{4}[/tex]
Conclusion
From the calculations above we can rightly say that the magnifications of the values above are
A has [tex]u=3\frac{f}{2}[/tex]
B has [tex]u=4\frac{f}{3}[/tex]
C has [tex]u=5\frac{f}{4}[/tex]
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