JEE Main 17 March 2021 Shift 1 Solved Paper

Given, thickness at the centre of plano-convex lens, t=3mm =3×10−3m
Diameter of plano-convex lens, d=6cm
∴ Radius of plano-convex lens, r=3cm=3×10−2m
Speed of light in lens material, v=2×108ms−1
Refractive index =‌

‌ Speed of light in air ‌

‌ Speed of light in medium ‌

µ=‌

3×108

2×108

=‌

3

2

=1.5
⇒‌‌µ=1.5We know that, R2=(R−t)2+r2
where, R is the radius of curvature of plano convex lens.
As, t is small, then t2 will be very very small, so it can be neglected.
⇒‌‌2Rt=r2
⇒‌‌R=r2/2t....(i)
From lens Maker's formula, we have
‌

1

f

=(µ−1)(‌

1

R1

−‌

1

R2

)
Here R1=R&R2=∞
⇒‌‌‌

1

f

=(µ−1)(‌

1

R

−‌

1

∞

)
⇒‌‌‌

1

f

=‌

µ−1

R

...(ii)
From Eqs. (i) and (ii), we get
‌

1

f

=‌

µ−1(2t)

r2

⇒f=‌

r2

(µ−1)2t

Putting the given values in above equation, we get
f=‌

(3×10−2)2

(1.5−1)2×3×10−3

⇒f=0.3m
f=30cm