Find the focal length of a meniscus lens with \(R_{1}=20 \mathrm{cm}\) and \(R_{2}=15 \mathrm{cm} .\) Assume that the index of refraction of the lens is \(1.5 .\)

Short Answer

Expert verified
Therefore, the focal length of the meniscus lens is approximately \( 120 cm \).

Step by step solution

01

Identifying the given

Identify the given in the problem. Here we are given: \( R_1 = 20 cm, R_2 = 15 cm \) which are the radii of curvature of the lens' surfaces, and \( n = 1.5 \) which is the refractive index of the lens material.
02

Applying Lensmaker's formula

The lensmaker's formula for a thin lens is given as: \( \frac{1}{f} = (n - 1)( \frac{1}{R_1} - \frac{1}{R_2} )\), where \( f \) is the focal length, \( n \) is the index of refraction and \( R_1 \) and \( R_2 \) are the radii of curvature of the lens' surfaces. Now we just insert the given values into the formula.
03

Solving for focal length

After inserting the given values into lensmaker's formula: \( \frac{1}{f} = (1.5 - 1)( \frac{1}{20 cm} - \frac{1}{15 cm} )\), it simplifies to : \( \frac{1}{f} = 0.5(cm^{-1}) * (0.05 cm - 0.0667 cm)\). Further simplifying gives: \( \frac{1}{f} = -0.5 * (-0.0167 cm) = 0.00835 cm^{-1}\). The focal length will be the reciprocal of this result.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.

Sign-up for free