How Do Sign Conventions Affect Calculations in Spherical Mirrors?

[alfredbester]

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.

 

[alfredbester]

Also.
An object is placed 5 cm from a thin equi-convex lens of focal length 4cm. Another thin equi-convex lens is placed on the same axis as the first and 9cm away from it, on the side further from the object. It has a focal length of 5cm.

(i) Find the position of the image. Is the image real

A: The image will be formed 3.475 cm to the right of lens 2 (therefore it's real ), or 17.475cm to the right of the object.

(ii) What are the magnifications of the intermediate and final image.

A: M(int) = -Si1/So1 = -(20cm)/(5cm) = -4
M(2) = -(Si2)/(So2) = -(55/16)/(-11) = 0.3125
M(tot) = M(int).M(2) = -1.25

Would appreciate somebody telling me it's all correct as well as pointing out any mistakes.

 

[alfredbester]

Bump. I take it nobody does optics then.

 

[nrqed]

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.

This looks right to me. the signs are ok. A quick check with a ray diagram agrees qualitatively with your results too. (it`s useful to know how to draw ray diagrams to check results like this).

Pat