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A spherical surface of radius of curvature R separates air (refractive index 1.0) from glass (refractive index 1.5). The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at a point O, and PO = OQ. The distance PO is equal to
- a)5R
- b)3R
- c)2R
- d)1.5R
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A spherical surface of radius of curvature R separates air (refractive...
The formula for spherical refracting surface is
Here u = –x, v = + x, R = + R, µ1 = 1, µ2 = 1.5
Most Upvoted Answer
A spherical surface of radius of curvature R separates air (refractive...
Given information:
- Radius of curvature of the spherical surface: R
- Refractive index of air: 1.0
- Refractive index of glass: 1.5
- Centre of curvature is in the glass
- Point object P placed in air has a real image Q in the glass
- Line PQ cuts the surface at a point O
- PO = OQ
To find: The distance PO
Explanation:
To solve this problem, we can use the lens formula:
1/f = (n2 - n1) * (1/R1 - 1/R2)
where,
f = focal length of the lens
n1 = refractive index of the medium on the object side (air)
n2 = refractive index of the medium on the image side (glass)
R1 = radius of curvature of the first surface of the lens
R2 = radius of curvature of the second surface of the lens
Since the spherical surface separates air and glass, we can consider it as a lens.
In this case, the object is in air and the image is in glass. Therefore, the lens formula becomes:
1/f = (1.5 - 1.0) * (1/R1 - 1/R2)
Since the centre of curvature is in the glass, the radius of curvature of the first surface (R1) is negative.
1/f = (1.5 - 1.0) * (1/R - 1/R2)
Since the point object P has a real image Q, the focal length of the lens (f) is positive.
Now, let's consider the point O. Since PO = OQ, the point O lies on the principal axis of the lens. Therefore, the distance PO is equal to the focal length of the lens.
PO = f
1/f = (1.5 - 1.0) * (1/R - 1/R2)
Simplifying this equation:
1/f = 0.5 * (1/R - 1/R2)
Since PO = f, we can write:
PO = 0.5 * (1/R - 1/R2)
Since the radius of curvature of the first surface (R1) is negative, we can write R1 = -R. Therefore:
PO = 0.5 * (1/R - 1/R2)
= 0.5 * (1/R + 1/R)
= 0.5 * (2/R)
= 1/R
Hence, the distance PO is equal to R.
Therefore, the correct answer is option 'A'.
- Radius of curvature of the spherical surface: R
- Refractive index of air: 1.0
- Refractive index of glass: 1.5
- Centre of curvature is in the glass
- Point object P placed in air has a real image Q in the glass
- Line PQ cuts the surface at a point O
- PO = OQ
To find: The distance PO
Explanation:
To solve this problem, we can use the lens formula:
1/f = (n2 - n1) * (1/R1 - 1/R2)
where,
f = focal length of the lens
n1 = refractive index of the medium on the object side (air)
n2 = refractive index of the medium on the image side (glass)
R1 = radius of curvature of the first surface of the lens
R2 = radius of curvature of the second surface of the lens
Since the spherical surface separates air and glass, we can consider it as a lens.
In this case, the object is in air and the image is in glass. Therefore, the lens formula becomes:
1/f = (1.5 - 1.0) * (1/R1 - 1/R2)
Since the centre of curvature is in the glass, the radius of curvature of the first surface (R1) is negative.
1/f = (1.5 - 1.0) * (1/R - 1/R2)
Since the point object P has a real image Q, the focal length of the lens (f) is positive.
Now, let's consider the point O. Since PO = OQ, the point O lies on the principal axis of the lens. Therefore, the distance PO is equal to the focal length of the lens.
PO = f
1/f = (1.5 - 1.0) * (1/R - 1/R2)
Simplifying this equation:
1/f = 0.5 * (1/R - 1/R2)
Since PO = f, we can write:
PO = 0.5 * (1/R - 1/R2)
Since the radius of curvature of the first surface (R1) is negative, we can write R1 = -R. Therefore:
PO = 0.5 * (1/R - 1/R2)
= 0.5 * (1/R + 1/R)
= 0.5 * (2/R)
= 1/R
Hence, the distance PO is equal to R.
Therefore, the correct answer is option 'A'.
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A spherical surface of radius of curvature R separates air (refractive index 1.0) from glass (refractive index 1.5). The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at a point O, and PO = OQ. The distance PO is equal toa)5Rb)3Rc)2Rd)1.5RCorrect answer is option 'A'. Can you explain this answer?
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A spherical surface of radius of curvature R separates air (refractive index 1.0) from glass (refractive index 1.5). The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at a point O, and PO = OQ. The distance PO is equal toa)5Rb)3Rc)2Rd)1.5RCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A spherical surface of radius of curvature R separates air (refractive index 1.0) from glass (refractive index 1.5). The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at a point O, and PO = OQ. The distance PO is equal toa)5Rb)3Rc)2Rd)1.5RCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A spherical surface of radius of curvature R separates air (refractive index 1.0) from glass (refractive index 1.5). The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at a point O, and PO = OQ. The distance PO is equal toa)5Rb)3Rc)2Rd)1.5RCorrect answer is option 'A'. Can you explain this answer?.
A spherical surface of radius of curvature R separates air (refractive index 1.0) from glass (refractive index 1.5). The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at a point O, and PO = OQ. The distance PO is equal toa)5Rb)3Rc)2Rd)1.5RCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A spherical surface of radius of curvature R separates air (refractive index 1.0) from glass (refractive index 1.5). The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at a point O, and PO = OQ. The distance PO is equal toa)5Rb)3Rc)2Rd)1.5RCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A spherical surface of radius of curvature R separates air (refractive index 1.0) from glass (refractive index 1.5). The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at a point O, and PO = OQ. The distance PO is equal toa)5Rb)3Rc)2Rd)1.5RCorrect answer is option 'A'. Can you explain this answer?.
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