Dispensing Optician Training HELP !!

Hi,
I am Nicola in the UK.
Stuck on a few questions.
Not looking to be spoon fed but some help would be great.
Here goes
A thin plus lens, made in glass of refractive index 1.50, has a focal length of +200mm and a concave surface radius of +125mm. What is the radius of the convex surface ?

If some one could talk me through it that would be great. :cheers:

First you need to find out what power you have. You know what the focal length is and you know that it is positive which means you are dealing with a plus lens.
$D = \frac{1}{f}_{m}$
This formula says the dioptric value is equal to the reciprocal of the focal length in meters.
200 mm is equal to 0.2 meters so:
$D = \frac{1}{0.2}\\ D = +5.00 D$

Now we need to find the power of our concave surface, we will use the radius of curvature formula.

$D = \frac{(n_{r}-n_{i})}{r_{m}}$

$n_{r}$ is the refractive index of the material that the ray is refracted in, which is air in this case.
$n_{i}$ is the refractive index of the material that the ray is incident in, which is 1.5 in this case.
Now put your values in the formula above.

$D = \frac{1-1.5}{0.125}\\ D = -4.00 D $

Since there is no thickness we can use a thin lens approximation formula called the nominal power formula, which state that the front power plus the back power is equal to the lens power.

$D_{n} = D_{1} + D_{2}$

We know the power is + 5.00 D and the back curve( $D_{2}$ )
-4.00 D

Now we can put these values into the formula above.
$+5.00 = D_{1} + (-4.00)\\ D_{1} = 5.00 + 4.00\\ D_{1} = +9.00 D$

The power of the front surface is +9.00, now we can put this in the radius formula. Keep in mind that the material that the ray is incident in for the front surface is air and it is refracted in the material.
$r_{m} = \frac{(1.5-1)}{9.00}\\ r_{m} = 0.056 $

I hope this helps. Good luck!

That was a big help, made everything totally clear.
Many Thanks Lensgrinder.

I might get in touch again if you dont mind.:cheers: