Question

Two thin lenses of power $+2.5D$ and $-0.5D$ are placed in contact. Find the power of the lens combination. Also, calculate the focal length of the combination.

Answer 1

Step 1: Given data,

Power of lens 1 is $P_1=+2.5D$

Power of lens 2 is $P_2=-0.5D$

Step 2: Finding the power of the lens combination,

According to power of lens combination formula,

$P=P_1+P_2$

Substituting the know values,

$$\begin{gathered} P=+2.5D+\left(-0.5D\right) \\ P=2.5-0.5 \\ P=+2D \end{gathered}$$

Thus, the power of the lens combination is $+2D$.

Step 3: Finding the focal length of lens 1,

According to the power of lens formula,

$P_1=\frac{1}{f_1}$

where $f_1$ is the focal length of the lens 1

Substituting the known values,

$$\begin{gathered} +2.5D=\frac{1}{f_1} \\ 2.5=\frac{1}{f_1} \\ {f}_1=\frac{1}{2.5} \\ {f}_1=0.4m \end{gathered}$$

Thus, the focal length of lens 1 is $0.4m$.

Step 4: Finding the focal length of lens 2,

According to the power of lens formula,

$P_2=\frac{1}{f_2}$

where $f_2$ is the focal length of the lens 1

Substituting the known values,

$$\begin{gathered} -0.5D=\frac{1}{f_2} \\ -0.5=\frac{1}{f_2} \\ {f}_2=-\frac{1}{0.5} \\ {f}_2=-2m \end{gathered}$$

Thus, the focal length of the lens 2 is $2m$.

Step 5: Finding the focal length of the lens combination,

According to the focal length of the lens combination formula,

$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$

where, $f$ is the focal length of lens combination.

Substituting the known values,

$$\begin{gathered} \frac{1}{f}=\frac{1}{0.4m}+\frac{1}{\left(-2m\right)} \\ \frac{1}{f}=\frac{1}{0.4}-\frac{1}{2} \\ \frac{1}{f}=2 \\ f=\frac{1}{2} \\ f=+0.5m \\ f=+50cm \end{gathered}$$

Thus, the focal length of the lens combination is $+50cm$.