Answers to Optics Practical 2

Optics Problems Week2.pdf

Question 1 §

Optics Practical 2 Question 1.excalidraw Focal length ($f$) was found by halving the radius of curvature (6cm): $f=\frac{R}{2}=3$.

The image distance ($d_i$) from the mirror can be found by measurement (4cm), or via calculation: $\frac{1}{d_i}=\frac{1}{f}-\frac{1}{d_o}$, where the distance from the object is 12cm. Thus, $\frac{1}{d_i}=\frac{1}{3}-\frac{1}{12}⟹d_i=4$.

The radius of the image can be calculated by solving the following equation for $r^{\prime }$: $\frac{r^{\prime }}{r}=\frac{d_i}{d_o}⟹\frac{r^{\prime }}{1.5}=\frac{4}{12}⟹r^{\prime }=\frac{1}{2}$.

Therefore, the final image is 4cm away from the mirror, inverted, and a third of the size (a circle with radius 0.5cm).

Question 2 §

Optics Practical 2 Question 2.excalidraw Glass has the approximate $n=1.5$, such that $R_1=10\text{cm}$, $R_2=13\text{cm}$. Thus, using the lens-maker’s equation:

$$\frac{1}{f}=(1.5-1)(\frac{1}{10}-\frac{1}{13})=\frac{3}{260}$$

$$\therefore f=\frac{260}{3}\approx 87\text{cm}$$

So, the focal length of the converging convex-concave glass lens is 87cm.

Question 3 §

Optics Practical 2 Question 3.excalidraw Again, for glass, the approximate value of $n=1.5$, with $R_1=10\text{cm}$ and $R_2=-15\text{cm}$.

Thus, using the lens-maker’s equation:

$$\frac{1}{f}=(1.5-1)(\frac{1}{10}-\frac{1}{-15})=\frac{1}{12}$$

$$\therefore f=12\text{cm}$$

If the light hit the other surface first, instead, then the following would be the equation:

$$\frac{1}{f}=(1.5-1)(\frac{1}{15}-\frac{1}{-10})=\frac{1}{12}$$

$$\therefore f=12\text{cm}$$

This is the same answer. So, the focal length of the biconvex glass lens is 12cm.

Question 4 §

Optics Practical 2 Question 4.excalidraw The focal length ($f$) is 12cm, where the distance from the object ($d_o$) is 4cm, thus (using the thin lens formula) we can find the image’s distance, $d_i$:

$$\frac{1}{12}=\frac{1}{4}+\frac{1}{d_i}⟹d_i=-6\text{cm}$$

Thus, the image is 6cm in front of the lens (and object) of the same orientation. Using this, we can find the magnification and thus the height of the image:

$$M=-\frac{-6}{4}=1.5⟹h^{\prime }=Mh=1.5\cdot 1.2=1.8\text{cm}$$

Therefore, the image is 6cm in front of the lens and object, the same orientation, and 1.5x larger than the object.