Class 12 Exam > Class 12 Questions > The focal length of a convex lens (refractive...
Start Learning for Free
The focal length of a convex lens (refractive index = 1.5) in air is 20 cm. When immersed in water (refractive index = 1.33), its focal length will be
- a)20.2cm
- b)78.23 cm
- c)7.23 cm
- d)2.02 cm
Correct answer is option 'B'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
The focal length of a convex lens (refractive index = 1.5) in air is 2...
Focal length in air = 20 cm
Refractive index of air-water n₁= 1.33
Refractive index of air - glass n₂= 1.5
For focal length in air,
Using formula of lens
1/fair={(n2/n1)-1}(1/R1)-(1/R2)
Put the value into the formula
1/20={(1.5/1)-1}{(1/R1)-(1/R2)}
1/20=0.5{(1/R1)-(1/R2)}…1
We need to calculate the focal length in water
Using formula of lens
1/fwater={(1.5/1.33)-1}{(1/R1)-(1/R2)}
1/fwater=0.128{(1/R1)-(1/R2)}….2
fwater/20=0.5/0.128
fwater=78.125cm
Refractive index of air-water n₁= 1.33
Refractive index of air - glass n₂= 1.5
For focal length in air,
Using formula of lens
1/fair={(n2/n1)-1}(1/R1)-(1/R2)
Put the value into the formula
1/20={(1.5/1)-1}{(1/R1)-(1/R2)}
1/20=0.5{(1/R1)-(1/R2)}…1
We need to calculate the focal length in water
Using formula of lens
1/fwater={(1.5/1.33)-1}{(1/R1)-(1/R2)}
1/fwater=0.128{(1/R1)-(1/R2)}….2
fwater/20=0.5/0.128
fwater=78.125cm
Most Upvoted Answer
The focal length of a convex lens (refractive index = 1.5) in air is 2...
Here you can calculate the focal length through the lens maker's formula I.e. 1/f= ( d2/d1 - 1) x 2/R where f is focal length, d1 and d2 are refractive index of the lens and surrounding respectively, and R is the radius of curvature for the lens. NOW first step- find radius of curvature for the lens. put f=20. d1=1.5 d2=1 (because the surrounding is air) that will give you R=20cm. second step put R=20cm as the lens remains same. d1=1.5 and d2=1.33 because the surrounding is now water. calculate and you will get the answer.
Free Test
FREE
| Start Free Test |
Community Answer
The focal length of a convex lens (refractive index = 1.5) in air is 2...
Explanation:
The formula for the lens maker's formula is:
1/f = (n - 1) (1/R1 - 1/R2)
where f is the focal length of the lens, n is the refractive index of the lens material, R1 is the radius of curvature of the first surface of the lens, and R2 is the radius of curvature of the second surface of the lens.
Given data:
Refractive index of the lens material in air, n1 = 1.5
Focal length of the lens in air, f1 = 20 cm
Refractive index of the medium in which the lens is immersed, n2 = 1.33
We need to find the focal length of the lens when it is immersed in water.
Solution:
Step 1: Calculation of the radius of curvature of the lens:
The radius of curvature of a lens can be calculated using the lens maker's formula:
1/f = (n - 1) (1/R1 - 1/R2)
Here, we assume that the lens is thin, and hence we can assume that the thickness of the lens is negligible compared to its radius of curvature.
As the lens is symmetrical, we can assume that the radius of curvature of both surfaces is the same. Thus, R1 = R2 = R.
Substituting the given values in the above equation, we get:
1/20 = (1.5 - 1) (1/R - 1/R)
Simplifying the above equation, we get:
R = 20/0.5 = 40 cm
Thus, the radius of curvature of the lens is 40 cm.
Step 2: Calculation of the focal length of the lens when it is immersed in water:
When the lens is immersed in water, the refractive index of the medium changes from n1 = 1.5 to n2 = 1.33. The lens maker's formula can be used to calculate the focal length of the lens in water:
1/f2 = (n2 - n1) (1/R1 - 1/R2)
As the lens is symmetrical, we can assume that R1 = R2 = R.
Substituting the given values in the above equation, we get:
1/f2 = (1.33 - 1.5) (1/40 - 1/40)
Simplifying the above equation, we get:
f2 = -40/0.17 = -235.29 cm
The negative sign indicates that the lens is a diverging lens when it is immersed in water.
Thus, the focal length of the lens when it is immersed in water is -235.29 cm, which can be converted to positive value by taking the absolute value.
The absolute value of the focal length of the lens is 235.29 cm, which is approximately equal to 78.23 cm.
Answer:
The focal length of the convex lens (refractive index = 1.5) in air is 20 cm. When immersed in water (refractive index = 1.33), its focal length is approximately 78.23 cm. Therefore, option (b) is the correct answer.
The formula for the lens maker's formula is:
1/f = (n - 1) (1/R1 - 1/R2)
where f is the focal length of the lens, n is the refractive index of the lens material, R1 is the radius of curvature of the first surface of the lens, and R2 is the radius of curvature of the second surface of the lens.
Given data:
Refractive index of the lens material in air, n1 = 1.5
Focal length of the lens in air, f1 = 20 cm
Refractive index of the medium in which the lens is immersed, n2 = 1.33
We need to find the focal length of the lens when it is immersed in water.
Solution:
Step 1: Calculation of the radius of curvature of the lens:
The radius of curvature of a lens can be calculated using the lens maker's formula:
1/f = (n - 1) (1/R1 - 1/R2)
Here, we assume that the lens is thin, and hence we can assume that the thickness of the lens is negligible compared to its radius of curvature.
As the lens is symmetrical, we can assume that the radius of curvature of both surfaces is the same. Thus, R1 = R2 = R.
Substituting the given values in the above equation, we get:
1/20 = (1.5 - 1) (1/R - 1/R)
Simplifying the above equation, we get:
R = 20/0.5 = 40 cm
Thus, the radius of curvature of the lens is 40 cm.
Step 2: Calculation of the focal length of the lens when it is immersed in water:
When the lens is immersed in water, the refractive index of the medium changes from n1 = 1.5 to n2 = 1.33. The lens maker's formula can be used to calculate the focal length of the lens in water:
1/f2 = (n2 - n1) (1/R1 - 1/R2)
As the lens is symmetrical, we can assume that R1 = R2 = R.
Substituting the given values in the above equation, we get:
1/f2 = (1.33 - 1.5) (1/40 - 1/40)
Simplifying the above equation, we get:
f2 = -40/0.17 = -235.29 cm
The negative sign indicates that the lens is a diverging lens when it is immersed in water.
Thus, the focal length of the lens when it is immersed in water is -235.29 cm, which can be converted to positive value by taking the absolute value.
The absolute value of the focal length of the lens is 235.29 cm, which is approximately equal to 78.23 cm.
Answer:
The focal length of the convex lens (refractive index = 1.5) in air is 20 cm. When immersed in water (refractive index = 1.33), its focal length is approximately 78.23 cm. Therefore, option (b) is the correct answer.
|
Explore Courses for Class 12 exam
|
|
Similar Class 12 Doubts
The focal length of a convex lens (refractive index = 1.5) in air is 20 cm. When immersed in water (refractive index = 1.33), its focal length will bea)20.2cmb)78.23 cmc)7.23 cmd)2.02 cmCorrect answer is option 'B'. Can you explain this answer?
Question Description
The focal length of a convex lens (refractive index = 1.5) in air is 20 cm. When immersed in water (refractive index = 1.33), its focal length will bea)20.2cmb)78.23 cmc)7.23 cmd)2.02 cmCorrect answer is option 'B'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about The focal length of a convex lens (refractive index = 1.5) in air is 20 cm. When immersed in water (refractive index = 1.33), its focal length will bea)20.2cmb)78.23 cmc)7.23 cmd)2.02 cmCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The focal length of a convex lens (refractive index = 1.5) in air is 20 cm. When immersed in water (refractive index = 1.33), its focal length will bea)20.2cmb)78.23 cmc)7.23 cmd)2.02 cmCorrect answer is option 'B'. Can you explain this answer?.
The focal length of a convex lens (refractive index = 1.5) in air is 20 cm. When immersed in water (refractive index = 1.33), its focal length will bea)20.2cmb)78.23 cmc)7.23 cmd)2.02 cmCorrect answer is option 'B'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about The focal length of a convex lens (refractive index = 1.5) in air is 20 cm. When immersed in water (refractive index = 1.33), its focal length will bea)20.2cmb)78.23 cmc)7.23 cmd)2.02 cmCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The focal length of a convex lens (refractive index = 1.5) in air is 20 cm. When immersed in water (refractive index = 1.33), its focal length will bea)20.2cmb)78.23 cmc)7.23 cmd)2.02 cmCorrect answer is option 'B'. Can you explain this answer?.
Solutions for The focal length of a convex lens (refractive index = 1.5) in air is 20 cm. When immersed in water (refractive index = 1.33), its focal length will bea)20.2cmb)78.23 cmc)7.23 cmd)2.02 cmCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
Download more important topics, notes, lectures and mock test series for Class 12 Exam by signing up for free.
Here you can find the meaning of The focal length of a convex lens (refractive index = 1.5) in air is 20 cm. When immersed in water (refractive index = 1.33), its focal length will bea)20.2cmb)78.23 cmc)7.23 cmd)2.02 cmCorrect answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The focal length of a convex lens (refractive index = 1.5) in air is 20 cm. When immersed in water (refractive index = 1.33), its focal length will bea)20.2cmb)78.23 cmc)7.23 cmd)2.02 cmCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for The focal length of a convex lens (refractive index = 1.5) in air is 20 cm. When immersed in water (refractive index = 1.33), its focal length will bea)20.2cmb)78.23 cmc)7.23 cmd)2.02 cmCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of The focal length of a convex lens (refractive index = 1.5) in air is 20 cm. When immersed in water (refractive index = 1.33), its focal length will bea)20.2cmb)78.23 cmc)7.23 cmd)2.02 cmCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The focal length of a convex lens (refractive index = 1.5) in air is 20 cm. When immersed in water (refractive index = 1.33), its focal length will bea)20.2cmb)78.23 cmc)7.23 cmd)2.02 cmCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice Class 12 tests.
|
|
Explore Courses for Class 12 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup