Question

18. A biconvex lens has radii of curvature 12cm and 18cm and focal length of 14.4cm. Find the
Value of refractive index of the material of material of lens.​

Answer 1

Answer:

From lens maker's formula,

f

1

=(μ−1)(

R

1

1

−

R

2

1

)

We have,

Refractive index, μ=1.5 and R

1

=20 cm and R

2

=30 cm

f

1

=(1.5−1)(

R

1

1

−

R

2

1

)

f

1

=(1.5−1)(

−20

1

−

−30

1

)

f

1

=(1.5−1)

20×30

−30+20

f=−120cm

Step-by-step explanation:

From lens maker's formula,

f

1

=(μ−1)(

R

1

1

−

R

2

1

)

We have,

Refractive index, μ=1.5 and R

1

=20 cm and R

2

=30 cm

f

1

=(1.5−1)(

R

1

1

−

R

2

1

)

f

1

=(1.5−1)(

−20

1

−

−30

1

)

f

1

=(1.5−1)

20×30

−30+20

f=−120cm

Answer 2

Answer:

48 is the answer please mark me as brainlist