JEE Advanced 2019 Paper 1

f=‌‌

R

2(n−1)

...........(1)
‌‌

1

f+df

=‌‌

1

f

+‌‌

dn

R

Clearly,if dn<0,R.H.S value is less than

1

f

∴df>0i.e (1) is correct
Option(2)
f=‌‌

R

2(1.5−1)

⇒R=20cm
‌‌

1

f+df

=‌‌

1

f

+‌‌

dn

R

‌‌

1

20+df

=‌‌

1

20

+‌‌

dn

20

‌‌

1

20(1+‌‌ df 20 )

=‌‌

1

20

(1+dn)
1−‌‌

df

20

=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
−‌‌

df

f

=(‌‌

dn

n

)‌‌

1

2(1−‌‌ 1 n )

As this is independent of R, (3) is correct
Option (4):|‌‌

df

f

|=|‌‌

dn

n

|‌‌

1

2(1−‌‌ 1 n )

For the range 1<n<2,2(1−‌‌

1

n

) is always greater than 1 i.e,|‌‌

df

f

|>|‌‌

dn

n

|
∴ (4) is wrong