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Consider a concave mirror that has a focal length f. In terms of f, determine the object distances that will produce a magnification of

A. -2
B. -3
C. -4


Sagot :

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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