A concave lens has a focal …
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Posted by Shubham Saxena ???? 3 years, 7 months ago
- 3 answers
Anuj Mittal 3 years, 7 months ago
F= - 10 CM
H1 =2.5 CM
U = -30CM
{tex}{ 1\over F } = {1\over v } -{ 1\over u}{/tex}
1/-10=1/v -1/-30
1/-10 +1/30 = 1/v
(-3+1)/30=1/v
-2/30 = 1/v
-1/15 =1/v
v = -15
position is on left side 15 cm away from the lens
{tex}{H2 \over H1 } = {v \over u}{/tex}
H2 = (-15*2.5)/-30
H2 = 1.25 cm
as it is positive so image is virtual and erect
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Haggal Kumar 3 years, 6 months ago
0Thank You