Question

A point object is kept 10 cm away from a
double convex lens of refractive index
1.5 and radii of curvature 10 cm and 8 cm
Find the focal length of the lens​

Answer 1

Answer:

Answer is given down

Explanation:

Hi,

Answer:

The location of the final image is 120 cm.

Explanation:

Given data:

In a double convex lens,

Object distance, u = -10 cm

Refractive index, n = 1.5

Radius of curvatures, R1 = 10 cm and R2 = -8 cm

Central thickness of the lens, t = 2 cm

To find: final image distance, “v” considering paraxial only

For thick convex lens, we have

Focal length, f is positive

R1 is positive

R2 is negative

Object distance “u” is negative and image distance “v” is positive

Image formed is real & inverted

We know, the lens maker’s formula is given as

\frac{1}{f}

f

1

= [n-1][\frac{1}{R1}

R1

1

– \frac{1}{R2}

R2

1

+ \frac{(n - 1) * t}{nR1R2}

nR1R2

(n−1)∗t

Or, \frac{1}{f}

f

1

= [1.5-1][(1/10) + (1/8) - {(1.5 - 1)*2 / (1.5 * 10 * 8)}]

Or, \frac{1}{f}

f

1

= [0.5] [0.1 + 0.125 – (1/120)]

Or, \frac{1}{f}

f

1

= [0.5] [0.21667] = 0.108335

Now, as per the general lens formula, we have

\frac{1}{f}

f

1

= \frac{1}{v}

v

1

– \frac{1}{u}

u

1

Or, 0.108335 = 1/v – 1/(-10)

Or, 0.108335 – 0.1 = 1/v

Or, \frac{1}{v}

v

1

= 0.008335

∴ v = 1/0.008335 = 119.97 cm ≈ 120cm

Hence, the location of the final image is 120 cm away from the lens considering paraxial only.

Hope this helps!!!!!