The image distance is 22.5 cm
Explanation:
We can solve the problem by using the mirror equation:
[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}[/tex]
where
f is the forcal length
p is the distance of the object from the mirror
q is the distance of the image from the mirror
For this problem, we have:
f = +15.0 cm (focal length is positive for a concave mirror)
p = 15.0 cm (distance of the object)
Therefore, solving for the image distance, we find:
[tex]\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{15}-\frac{1}{45}=\frac{2}{45}\\\rightarrow q=\frac{45}{2}=22.5 cm[/tex]
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The image distance will be "22.5 cm".
According to the question,
Focal length, f = 15.0 cm
Distance of object, p = 15.0 cm
Let,
The image distance be "q".
By using the mirror equation,
→ [tex]\frac{1}{f}[/tex] = [tex]\frac{1}{p} + \frac{1}{q}[/tex]
and,
→ [tex]\frac{1}{q}[/tex] = [tex]\frac{1}{f} - \frac{1}{p}[/tex]
By substituting the values, we get
= [tex]\frac{1}{15} - \frac{1}{45}[/tex]
By taking L.C.M, we get
= [tex]\frac{3-1}{45}[/tex]
[tex]\frac{1}{q}[/tex] = [tex]\frac{2}{45}[/tex]
By applying cross-multiplication,
q = [tex]\frac{45}{2}[/tex]
= 22.5 cm
Thus the above response is correct.
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