Lens Calculations (AQA A Level Physics): Revision Note

A diverging lens has a focal length of −1.5 cm.

Calculate the power of the lens.

[2]

Answer:

Step 1: Write the known values

• Focal length, $f=-1.5\mathrm{cm}$

Step 2: Convert the focal length into m

• $1\mathrm{m}=100\mathrm{cm}$

• So, $-1.5\mathrm{cm}=\frac{-1.5}{100}=-0.015\mathrm{m}$

Step 3: Recall the equation for power and focal length

$P=\frac{1}{f}$

Step 4: Substitute for focal length into the equation

$P=\frac{1}{-0.015}$ [1 mark]

Step 5: Calculate the power of the lens, P

$P=-66.6\mathrm{D}$ [1 mark]

A lens is set up between a candle and a screen. All the equipment is arranged perpendicular to the desk. The image of the candle is in focus when the screen is positioned 60 cm from the lens.

Determine the power of the lens.

[2]

Answer:

Step 1: Write the known quantities in S.I. units

• The image is in focus, so the screen is at the focal point

• Focal length, $f=60\mathrm{cm}=0.6\mathrm{m}$ [1 mark]

Step 2: Write the equation for power and substitute the values

$P=\frac{1}{f}=\frac{1}{0.6}=1.7\mathrm{D}\left(2\mathrm{s}.\mathrm{f}.\right)$ [1 mark]

The explanations above relate to life in a common-sense way. Stronger, more powerful reading glasses are used by people who have the weakest eyesight. They need lenses to do more of the focusing for them. 

A student investigates the focal length of a thin lens by using it to project an image onto a screen.

The object is set 50.0 cm from the centre of the lens, and the screen is moved back and forth until the inverted image is sharp. This position is found to be 75.0 cm from the lens.

Determine the focal length of the lens.

[2]

Answer:

Step 1: Write the known values

• Distance from object to lens, $u=50.0\mathrm{cm}$

• Distance from image to lens, $v=75\mathrm{cm}$

Step 2: Write the equation and substitute in the values

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

$\frac{1}{f}=\frac{1}{50}+\frac{1}{75}=\frac{1}{30}$ [1 mark]

$f=30\mathrm{cm}$ [1 mark]

It is easy to forget the last step in a calculation like this one. Remember that you are calculating $\frac{1}{f}$ not $f$, and that you need to take the reciprocal of your answer.

A magnifying glass has a focal length of 15 cm. It is held 5 cm away from a component which is being examined. 

Determine the magnification of the image.

[2]

Answer:

Step 1: Write the known values

• Focal length, $f=15\mathrm{cm}$

• Distance between object and lens, $u=5\mathrm{cm}$

Step 2: Use the lens formula and rearrange to make v the subject

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

$v={\left(\frac{1}{f}-\frac{1}{u}\right)}^{-1}={\left(\frac{1}{15}-\frac{1}{5}\right)}^{-1}=-7.5\mathrm{cm}$ [1 mark]

• The negative sign indicates a virtual image (expected for a magnifying glass) and is ignored for the next step

Step 3: Use the magnification formula to find the magnification of the image

$m=\frac{v}{u}=\frac{7.5}{5}=1.5$ [1 mark]

A person sees an image from a magnifying glass.

Calculate the magnification of this image. Clearly show your working on the diagram.

[3]

Answer:

Step 1: Measure the height of the object from the scale

• The object is 10 cm [1 mark]

Step 2: Measure the height of image from the scale

• The image is 20 cm [1 mark]

Step 3: Substitute values into the magnification equation

$\mathrm{Magnification}=\frac{\mathrm{Image}\mathrm{height}}{\mathrm{Object}\mathrm{height}}=\frac{20}{10}=2$ [1 mark]

The most common mistake with magnification calculations is to get the formula upside down.

Do a 'sanity check' by looking at the answer to make sure that magnified objects have got bigger ($m>1$) and diminished ones smaller ($m<1$).

Since we are working with ratios (so the units get cancelled out) this is one of those rare times when you don't need to convert everything to SI units, but do check that your units are all the same - for example all distances in cm.