How Do Sign Conventions Affect Focal Length Calculations?

[Noreturn]

Homework Statement

1) A double convex thin lens has a radius of 25 cm for the first surface and a radius of 40 cm for the second surface. If the glass has a refractive index of 1.55 the focal length is a) 28 cm, b) 9.93 cm, c) 121 cm, or d) 63 cm.

2) An object is placed 16 cm in front of a thin concave lens of 10 cm focal length, therefore the image distance

Homework Equations

1/f=(n-1)((1/R1)-(1/R2))

1/f=(1/s)+(1/s^1)

The Attempt at a Solution

f=((1.55-1)((1/25)-(1/40)))^-1121cm

s^1=((1/10)-(1/16))^-1
26.67

However, they have the answer as A) 28cm and -6.15

 

[TSny]

Review the various sign conventions. You have one sign error in your calculation of f and one sign error in your calculation of s.

 

[RedDelicious]

In optics, you have to be especially cautious of your signs.

1) Check your signs for R1 and R2 for a double convex lens.

2) Check your focal length for a diverging lens.

 

[Noreturn]

Where did I mess up, It looks like I followed the equation that was given to us. I tried adding instead of subtracting and still got the wrong answer.

 

[TSny]

When is R negative?

When is f negative?

 

[Noreturn]

I still was not able to get this answer correctly, or answer when it's negative :(

Usually it's negative for diverging lens correct?

 

[TSny]

Please show your calculation with all of the individual steps and numbers that you used. Otherwise, we can't identify where you are making an error.

 

[Noreturn]

f=((1.55-1)((1/25)+(1/40)))^-1
f= ( (.55) (.065) )^-1

f= 28cm

So I did figure out 2:
s= ((1/10)+(1/16))^-1

Then the answer is negative on part 2 why? Then on both of them we have to end up adding instead of subtracting why is that?

 

[TSny]

Then on both of them we have to end up adding instead of subtracting why is that?

As mentioned earlier, the reason has to do with sign conventions for R and f. These sign conventions should have been stated in your textbook or in your lectures.

In part (a) you are given that the radius of curvatures of the first and second surfaces are 25 cm and 40 cm, respectively. When using the formula that contains R₁ and R₂, you need to use the sign conventions in order to decide if R₁ = +25 cm or R₁ = -25 cm. Likewise, the sign conventions will tell you whether to use R₂ = +40 cm or R₂ = -40 cm.

Similarly, for part (b), you need to know the sign convention for when f is positive and when f is negative. When the problem states that the focal length is 10 cm, they are just giving you the magnitude (absolute value) of the focal length. They are leaving it to you to decide if f = + 10 cm or f = -10 cm when using the formula.