Answer: 19.80 cm
Explanation:
Given
focal length [tex]f=6.14\ cm[/tex] (as focal length is positive, it is converging lens)
Image distance [tex]v=1.5f[/tex]
[tex]v=1.5\times 6.14\\v=9.21\ cm[/tex]
using lens formula
[tex]\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}[/tex]
insert values
[tex]\dfrac{1}{u}=\dfrac{1}{6.14}-\dfrac{1}{8.9}\\\\\dfrac{1}{u}=0.1628-0.1123\\\\\dfrac{1}{u}=0.0505\\\\u=19.80\ cm[/tex]
Thus, the distance of the object is 19.80 cm
Given:
Now,
The image distance will be:
→ [tex]v = 1.5 f[/tex]
[tex]= 1.5\times 6.14[/tex]
[tex]= 9.21 \ cm[/tex]
By using the lens formula, we get
→ [tex]\frac{1}{v} + \frac{1}{u} = \frac{1}{f}[/tex]
By putting the values, we get
→ [tex]\frac{1}{u} = \frac{1}{6.14} - \frac{1}{8.9}[/tex]
[tex]\frac{1}{u} = 0.1628-0.1123[/tex]
[tex]\frac{1}{u} = 0.0505[/tex]
[tex]u = 19.80 \ cm[/tex]
Thus the answer above is correct.
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