TS EAMCET 5-Aug-2021 Shift 2 Question Paper

Given,
Object distance from biconvex lens, u=−20cm
Radii of curvature, R=5cm
Refractive index, µ=1.5
As we know that,
‌

1

f

=(µ−1)(‌

1

R1

−‌

1

R2

)
where, f is focal length and R1,R2 are radii of curvatures of both surfaces.
‌ i.e. ‌‌R1=R2=R
∴‌‌

1

f

=(1.5−1)(‌

1

R

+‌

1

R

)=‌

1

2

⋅‌

2

R

=‌

1

R

∴‌f=R=5cm
By using lens formula,
‌

1

f

=‌

1

v

−‌

1

u

where, v is image distance.
∴‌‌‌

1

v

‌‌=‌

1

f

+‌

1

u

‌‌=‌

1

5

−‌

1

20

=‌

3

20

⇒‌‌v‌‌=20∕3cm