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# the focal length of a biconvex lens with refractive index 1.5 and radius of curvature r on both sides is

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## The focal length of a biconvex lens is 20 cm and its refractive index is 1.5. If the radii of curvatures of two surfaces of lens are in the ratio 1 : 2, then the larger radius of curvature is :

Click here👆to get an answer to your question ✍️ The focal length of a biconvex lens is 20 cm and its refractive index is 1.5. If the radii of curvatures of two surfaces of lens are in the ratio 1 : 2, then the larger radius of curvature is : Question

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## 30 cm

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Correct option is D)

From lens maker's formula,

f 1 ​ =(μ−1)( R 1 ​ 1 ​ − R 2 ​ 1 ​ ) 20 1 ​ =(1.5−1)( R 1 ​ − (−2R) 1 ​ ) 20 1 ​ = 2 1 ​ × 2R 3 ​ R=15cm

∴  length of  radius =(2R)=2×15 cm=30 cm

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## The focal length of a biconvex lens is 20 cm and its refractive index is 1.5. If the radii of curvatures of two surfaces of lens are in the ratio 1:2, then the larger radius of curvature is (in cm)

1/f=(mu-1)(1/R(1) - 1/R(2))The focal length of a biconvex lens is 20 cm and its refractive index is 1.5. If the radii of curvatures of two surfaces of lens are in the ratio 1:2, then the larger radius of curvature is (in cm) Home English Class 12 Physics Chapter

Ray Optics And Optical Instrauments

The focal length of a biconvex...

## The focal length of a biconvex lens is 20 cm and its refractive index is 1.5. If the radii of curvatures of two surfaces of lens are in the ratio 1:2, then the larger radius of curvature is (in cm)

Updated On: 17-04-2022

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`1/f=(mu-1)(1/R_(1) - 1/R_(2))`

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## Very Important Questions

एक गोलाकार आवेशित चालक का आवेश घनत्व `sigma` है। इसके पृष्ठ पर वैद्युत क्षेत्र E है। यदि गोले की त्रिज्या उसका आवेश घनत्व अपरिवर्तित रखते हुए दोगुनी कर दी जाए, तो नए गोले के पृष्ठ पर वैद्युत क्षेत्र होगा

दो बिन्दु आवेश `1muC` तथा `-1muC` एक दूसरे से `100Å` की दूरी द्वारा अलग है। द्विध्रुव के लम्ब द्विभाजक पर द्विध्रुव से 10 सेमी की दूरी पर एक बिन्दु P है। P पर वैद्युत क्षेत्र होगा

दो बिन्दु आवेशों के बीच की दूरी 10% बढ़ाते है, इनके बीच आकर्षण बल

दो बिन्दु आवेश `+2muC` तथा `+6muC` एक-दूसरे को 12 न्यूटन बल से प्रतिकर्षण करते है। यदि प्रत्येक `-4muC` का अतिरिक्त आवेश दिया जाता है, तब नया बल क्या होगा?

जल `(H_(2)O)` के एक उदासीन अणु का वाष्प अवस्था में वैद्युत द्विध्रुव आघूर्ण `6xx10^(-30)` सेमी है। यदि अणु को `1.5xx10^(4)` न्यूटन/कूलॉम वैद्युत क्षेत्र में रखा जाता है, तो क्षेत्र पर महत्तम बलाघूर्ण लगा सकता है, लगभग

यदि दो आवेश 5 सेमी की दूरी पर स्थित है। यदि इनके बीच एक ब्रास प्लेट रखी जाती है, तो दोनों आवेशों के बीच बल होगा

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## Why is the focal length of a convex lens equal to the radius of curvature when the refractive index is 1.5?

Answer: That isn’t necessarily true. The focal length of an optical surface is given by this equation: f=R/(n'-n) , where R is the radius of curvature; n’ is index of refraction for the glass; and n’ is the index of refraction for the incident light ray’s medium. f is only equal to R when n'-n=1. Why is the focal length of a convex lens equal to the radius of curvature when the refractive index is 1.5?

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, studied at Northmont High School

That isn’t necessarily true.

The focal length of an optical surface is given by this equation:

f=R/( n ′ −n) f=R/(n′−n)

, where R is the radius of curvature; n’ is index of refraction for the glass; and n’ is the index of refraction for the incident light ray’s medium. f is only equal to R when

n ′ −n=1. n′−n=1.

, studied Bachelor of Science in Computer Science & Photography at University of Delhi (2022)

Updated 1 year ago · Author has 90 answers and 74.8K answer views

Why is the focal length of a convex lens always positive?

OK, many answers given here on the basis of image formation on the right side of the optical center and only sign convention are WRONG.

Think about this, focal length is intrinsic property of a lens, and should not depend on how far the object or image is, though which side of lens they are on is important, which you’ll see why later in this answer.

ALSO, FOCAL LENGTH OF A CONVEX OF A CONVEX LENS IS NOT ALWAYS POSITIVE, ESPECIALLY IF THE OBJECT IS ON THE RIGHT OF THE LENS .

, former Structural Engineer and Project Manager (1980-2010)

The radius of curvature of the double convex lens are 10cm & 15cm. If the focal length is 12cm, what is the refractive index? Forge your way through the ages. Your empire awaits in the award-winning strategy game.

, retired teacher (1977-present)

Under what conditions will the radius of the curvature of a convex lens be equal to its focal length?

When the refractive index is equal to 1.5, the focal length of a convex lens is equal to its radius of curvatures.

, studied Physics & Chemistry at The University of Alabama in Huntsville (1976)

How can a lens have two different radii of curvature but the same focal length? I thought the focal length is half of the radius of curvature, or is it only for curved mirrors and not lenses?

Originally Answered: How can a lens have two different radii of curvature but the same focal length i thought focal length is half radius of curvature or is it only for curved mirrors only and not lenses?

A lens is a little more complicated than a mirror.

A mirror behaves mathematically as if the index were -1 and it only has one relevant surface.

A lens has two surfaces, distance between them, and a refractive index, and all of those properties affect the focal length.

The formula for the focal length of a lens in a vacuum is given by:

1 f =(n−1)[ 1 R 1 − 1 R 2 + (n−1)d n R 1 R 2 ]

1f=(n−1)[1R1−1R2+(n−1)dnR1R2]

where

f is the focal length

R 1 R1

is the radius of curvature of the first surface

R 2 R2

is the radius of curvature of the second surface

n is the index of refraction of the lens

d is Related Answer Chinmay Gaidhani

What is the relation between refractive index and focal length of a lens?

1/f = (r.i -1)(1/R1-1/R2)

R1 - radius of curvature of 1st surface

R2 - radius of curvature of 2 nd surface

r.i - Refractive Index

f - focal length Related Answer Soutrik Das

, Made some appscript as a freelancer

In a plano-convex lens the radius of curvature of the convex surface is 10 cm and the focal length of the lens is 30 cm. What will be the refractive index of the material of the lens?

Lens Makers formula says that

1 f =(μ−1)( 1 R 1 − 1 R 2 ) 1f=(μ−1)(1R1−1R2) So putting f=30cm f=30cm , R 1 =∞ R1=∞ and R 2 =−10cm R2=−10cm we get that μ= 4 3 μ=43

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