Optics

From the thin lens formula:

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

Multiplying through by $v$:

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

Since magnification $m = \frac{v}{u}$, this gives:

$$1 - m = \frac{v}{f} \implies m = 1 - \frac{v}{f}$$

Taking the magnitude/slope of the linear plot of $m$ versus $v$:

$$\left|\frac{dm}{dv}\right| = \frac{1}{f}$$

From the given graph:

• Vertical change (change in magnification, $\Delta m$) = $c$
• Horizontal change (change in image distance, $\Delta v$) = $b$

Analysis of light traveling from optically denser ($\mu_1$) to rarer ($\mu_2 \lt \mu_1$) medium:

• Statement A is incorrect: Energy of photons is conserved ($E = h f$).
• Statement B is correct: Frequency is determined solely by the source and does not change across boundaries.
• Statement C is incorrect: There is no $90^\circ$ phase change during normal refraction.
• Statement D is correct: Speed of light increases in rarer medium: $v = c / \mu_2 \gt c / \mu_1$.
• Statement E is incorrect: Wavelength increases in rarer medium: $\lambda = v / f$.

By the Lens Maker's formula:

$$\frac{1}{f} = (\mu - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

For a plano-convex lens with convex surface facing light:

• $R_1 = +20\,\text{cm}$
• $R_2 = \infty$ (plane surface)
• $\mu = 1.5$

Substituting the values:

$$\frac{1}{f} = (1.5 - 1) \left(\frac{1}{20} - \frac{1}{\infty}\right) = 0.5 \times \frac{1}{20} = \frac{1}{40}\quad\implies\quad f = +40\,\text{cm}$$