=1+dn⇒df=−20dn =−20×10−3cm ∴|df|=0.02cm (2) is correct Option(3): If convex surfaces are replaced with concave surfaces, sign of R will change
‌‌
1
f(1+‌‌
df
f
)
=‌‌
1
f
+‌‌
dn
R
=‌‌
1
f
+‌‌
dn
2f(n−1)
(
f=‌‌
R
2(n−1)
‌‌
1
R
=‌‌
1
2f(n−1)
) ‌‌
1−‌‌
df
f
f
=‌‌
1
f
[1+‌‌
dn
2n(1−‌‌
1
n
)
]
−‌‌
df
f
=(‌‌
dn
n
)‌‌
1
2(1−‌‌
1
n
)
As this is independent of R, (3) is correct Option (4):|‌‌