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Hi,

Doing an optics course I'm getting a bit confused about sign the conventions at time, can somebody check my answers here?

A spherical concave mirror of radius, R= 100mm has a real object placed at an axial distance So from its vertex.

Q1 Obtain the focal length of the mirror

1/ f = -2/R => f = -R/2 = -(-100mm)/2 = 50 mm right?

Q2 If an object is placed at a distance (i) Object distance, So = 200mm and then (ii) So = 20mm find the image positions, whether they are real or virtual and there magnifications.

(i) 1/Si = 1/f - 1/So = 1/50 - 1/200 = 150/10000 mm^-1. Si = image distance

M = -Si/So = -(10000/150 mm) / (200 mm) = -1/3

So the image is real, 67mm to the right of the mirror and is inverted and minified by a factor of 1/3.

(ii) 1/Si = 1/f - 1/So = 1/50 - 1/20 = - 30/1000 mm^-1

M = -Si/So = -(-1000/30 mm) / (20 mm) = 5/3

So the image is virtual, 33mm to the left of the mirror and is erect and magnified by a factor of 5/3.

Doing an optics course I'm getting a bit confused about sign the conventions at time, can somebody check my answers here?

A spherical concave mirror of radius, R= 100mm has a real object placed at an axial distance So from its vertex.

Q1 Obtain the focal length of the mirror

1/ f = -2/R => f = -R/2 = -(-100mm)/2 = 50 mm right?

Q2 If an object is placed at a distance (i) Object distance, So = 200mm and then (ii) So = 20mm find the image positions, whether they are real or virtual and there magnifications.

(i) 1/Si = 1/f - 1/So = 1/50 - 1/200 = 150/10000 mm^-1. Si = image distance

M = -Si/So = -(10000/150 mm) / (200 mm) = -1/3

So the image is real, 67mm to the right of the mirror and is inverted and minified by a factor of 1/3.

(ii) 1/Si = 1/f - 1/So = 1/50 - 1/20 = - 30/1000 mm^-1

M = -Si/So = -(-1000/30 mm) / (20 mm) = 5/3

So the image is virtual, 33mm to the left of the mirror and is erect and magnified by a factor of 5/3.

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