Chemistry, asked by nehabisht89, 11 months ago

Relation between the radius of curvature and focus in convex mirror​

Answers

Answered by misnaminglover
1

The focal length (f) of a lens is the distance between the center of the lens and the point at which the reflected light, of a beam of light travelling parallel to the center line, meets the center line (principal axis).

The radius of curvature (r) is the radius of the lens that forms a complete sphere.

The red line represents the light hitting the lens (AB) and reflecting off of the lens (BF). We know that the green line, RB, which represents the radius line of the circle, bisects the angle ABF because it is always at right angles to the lens. Therefore angles ABR and RBF are equal. We also know because of the rule of alternate angles that angle ABR is equal to angle BRF. Therefore, triangle BRF is an isosceles triangle. Consequently the lines BF and RF are equal.

We also know that BF and FC are approximately equal for lens that are small.

Therefore:

RF=BF=FC and

RC=RF+FC=FC+FC=2FC or

r=2f

Therefore, the radius of curvature is twice the focal length.


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Answered by Anonymous
9

Answer:

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Explanation:

The focal length (f) of a lens is the distance between the center of the lens and the point at which the reflected light, of a beam of light travelling parallel to the center line, meets the center line (principal axis).

The radius of curvature (r) is the radius of the lens that forms a complete sphere.

Looking at the attached diagram we can say the following:

The red line represents the light hitting the lens (AB) and reflecting off of the lens (BF). We know that the green line, RB, which represents the radius line of the circle, bisects the angle ABF because it is always at right angles to the lens. Therefore angles ABR and RBF are equal. We also know because of the rule of alternate angles that angle ABR is equal to angle BRF. Therefore, triangle BRF is an isosceles triangle. Consequently the lines BF and RF are equal.

We also know that BF and FC are approximately equal for lens that are small.

Therefore:

RF=BF=FC and

RC=RF+FC=FC+FC=2FC or

r=2f

Therefore, the radius of curvature is twice the focal length...

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