Q.To ensure almost 100 per cent transmittivity, photographic lenses are often coated with a thin layer of dielectric material. The refractive index of this material is intermediated between that of air and glass (which makes the optical element of the lens). A typically used dielectric film is $\text{MgF}_2$ ($n = 1.38$). What should the thickness of the film be so that at the center of the visible speetrum ($5500\ \text{\AA}$) there is maximum transmission.

Concept and Intuition

A lens's focal power depends on the relative refractive index of the lens material to its surrounding medium, not the lens's refractive index alone. If a lens is placed in a medium with a higher refractive index than the lens itself (here, lens $\mu=1.5$ vs liquid $\mu=2$), the relative refractive index becomes less than 1, and a convex lens actually behaves like a diverging lens — this is the classic 'lens becomes invisible/inverts behaviour' phenomenon.

Step-by-Step Solution

1. Lens maker's formula: $\dfrac{1}{f} = \left(\dfrac{\mu_{lens}}{\mu_{medium}} - 1\right)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$.
2. In air ($\mu_{medium}=1$): $P_{air} = \dfrac{1}{f_{air}} = (\mu_{lens}-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$.
3. Given $P_{air}=5$ D and $\mu_{lens}=1.5$: $5 = (1.5-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right) \Rightarrow \left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right) = \dfrac{5}{0.5}=10$. …

Concept and Intuition

A lens's focal length depends not just on its own refractive index but on the relative refractive index between the lens material and whatever medium surrounds it. Immersing the same lens in two different liquids changes the relative index, and hence the focal length, even though the lens's shape (its radii of curvature) stays the same.

Step-by-Step Solution

1. Lens maker's formula (in a medium): $\dfrac1f = \left(\dfrac{n_{lens}}{n_{medium}}-1\right)k$, where $k=\dfrac1{R_1}-\dfrac1{R_2}$ is fixed by the lens's shape.
2. In liquid 1 ($n=1.2$): $\dfrac1{f_1} = \left(\dfrac{1.5}{1.2}-1\right)k = (1.25-1)k = 0.25k$.
3. In liquid 2 ($n=1.25$): $\dfrac1{f_2} = \left(\dfrac{1.5}{1.25}-1\right)k = (1.2-1)k = 0.2k$. …

Concept and Intuition

A lens's power comes not from its own refractive index alone, but from the relative refractive index between the lens material and the surrounding medium (lensmaker's equation: $\frac1f = \left(\frac{n_{lens}}{n_{medium}}-1\right)\left(\frac1{R_1}-\frac1{R_2}\right)$). Surrounding a glass lens with water (whose refractive index is much closer to glass's than air's is) sharply reduces this relative index difference, weakening the lens and increasing its focal length. If the focal length grows enough that the (fixed) object distance becomes smaller than the new focal length, the lens can no longer converge the rays to a real image at all.

Step-by-Step Solution

1. Lensmaker's equation: $\dfrac1f = \left(\dfrac{n_{lens}}{n_{medium}}-1\right)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)$. The geometric factor $\left(\frac1{R_1}-\frac1{R_2}\right)$ is fixed by the lens shape and doesn't change.
2. In air ($n_{medium}=1$): relative-index factor $= 1.5 - 1 = 0.5$.
3. In water ($n_{medium} = 4/3$): relative-index factor $= \dfrac{1.5}{4/3} - 1 = 1.125 - 1 = 0.125$.
4. Since $\frac1f \propto$ this factor, $\dfrac{f_{water}}{f_{air}} = \dfrac{0.5}{0.125} = 4$, so $f_{water} = 4f_{air}$ — the focal length quadruples.
5. In the original air setup, the object was at $2f_{air}$ from the lens (as shown in the figure), forming a real image at $2f_{air}$ on the screen side.
6. In water, the same physical object distance is still $2f_{air}$, but the new focal length is $f_{water}=4f_{air}$. Since $2f_{air} < 4f_{air}$, the object now lies inside the (new, larger) focal length of the lens. …

Concept and Intuition

A lens only bends light because its refractive index differs from that of its surroundings. When a lens of index $n$ is immersed in a medium of index $n_m$, its effective 'relative' index becomes $n/n_m$, and its power scales accordingly: $\dfrac1f = \left(\dfrac{n}{n_m}-1\right)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)$. If the liquid's index is closer to the lens's own index, the lens becomes 'weaker' (larger focal length) — which is exactly what a longer $f_{liquid}$ tells us here.

Step-by-Step Solution

1. Let $K = \dfrac1{R_1}-\dfrac1{R_2}$ (same for both cases, since the lens shape doesn't change).
2. In air: $\dfrac1{f_{air}} = (n-1)K = (1.5-1)K = 0.5K$.
3. In liquid: $\dfrac1{f_{liq}} = \left(\dfrac{n}{n_{liq}}-1\right)K$.
4. Given $\dfrac{f_{air}}{f_{liq}} = \dfrac12$, so $\dfrac{f_{liq}}{f_{air}}=2$, i.e. $\dfrac{1/f_{air}}{1/f_{liq}} = 2$. …

Concept and Intuition

A converging lens can flip to a diverging one when placed in a medium whose refractive index exceeds the lens material's own index — because then the "relative" refractive index term $(n_{lens}/n_{medium}-1)$ becomes negative even though the lens shape is unchanged.

Step-by-Step Solution

1. In air: $\dfrac{1}{f_{air}} = (n_L-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$, with $f_{air}=0.2$ m, $n_L=1.5$:

2. In liquid, the lens behaves as a diverging lens of power $1$ D, so $f_{liq} = -1$ m. …

Concept and Intuition

A lens's focal length depends on the relative refractive index between the lens material and the surrounding medium: $\frac1f = \left(\frac{n_{lens}}{n_{medium}}-1\right)\left(\frac1{R_1}-\frac1{R_2}\right)$. Immersing a lens in a medium with a refractive index closer to the lens's own index reduces this relative index, which weakens the lens (increases its focal length) — exactly what's described here.

Step-by-Step Solution

1. Let $n$ be the lens's refractive index; the liquid's index is $n_{liq}=0.8n$.
2. In air: $\dfrac{1}{f_{air}} = (n-1)K$, where $K=\dfrac1{R_1}-\dfrac1{R_2}$.
3. In liquid: $\dfrac{1}{f_{liq}} = \left(\dfrac{n}{0.8n}-1\right)K = \left(1.25-1\right)K = 0.25K$. …

Concept and Intuition

When light passes through several parallel-faced media (air → oil → water, with the two interfaces parallel to each other), Snell's law applied at each interface can be chained together, and the intermediate refractive index always cancels: $n_1\sin\theta_1=n_2\sin\theta_2=n_3\sin\theta_3$ reduces to $n_1\sin\theta_1=n_3\sin\theta_3$ directly. So the oil's refractive index is a distractor here.

Step-by-Step Solution

1. Apply Snell's law directly from air to water (skipping oil since faces are parallel): $1\times\sin45^\circ=1.33\times\sin\theta_w$.
2. $\sin45^\circ=\dfrac{\sqrt2}{2}\approx0.7071$.
3. $\sin\theta_w=\dfrac{0.7071}{4/3}=\dfrac{3\times0.7071}{4}=\dfrac{3\sqrt2}{8}$.
4. Note $\left(\dfrac{3\sqrt2}{8}\right)^2=\dfrac{9\times2}{64}=\dfrac{18}{64}=\dfrac{9}{32}$, so $\sin\theta_w=\sqrt{9/32}$.
5. $\theta_w=\sin^{-1}\sqrt{9/32}$.

Common Mistakes …

Concept and Intuition

When you view an object at the bottom of a medium from air, refraction makes it appear closer to the surface than it really is: apparent depth $= \dfrac{\text{real depth}}{\mu}$. A microscope focused on the needle must be positioned exactly at this apparent-depth distance above the true bottom. If the medium (and hence $\mu$) changes while the real depth stays the same, the apparent depth changes, and the microscope has to be re-positioned by the difference between the two apparent depths.

Step-by-Step Solution

1. Real depth of tank, $h = 12$ cm.
2. With water: apparent depth $= 9$ cm $\Rightarrow \mu_{water} = h/9 = 12/9 = 4/3$ (consistent with water's known refractive index).
3. Liquid is replaced (same real depth $h=12$ cm), $\mu_{liquid} = 1.5$.
4. New apparent depth $= h/\mu_{liquid} = 12/1.5 = 8$ cm.
5. The needle's image has shifted from "9 cm below the surface" to "8 cm below the surface" — i.e. it appears to rise by $9-8=1$ cm. …

Concept and Intuition

The lensmaker's equation, $\dfrac{1}{f} = \left(\dfrac{n_{lens}}{n_{medium}}-1\right)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$, shows that a lens's power depends on the refractive index of the lens relative to its surroundings, not on the lens index alone. The geometric term $\left(\dfrac1{R_1}-\dfrac1{R_2}\right)$ never changes (same physical lens), so power in any medium can be found by scaling the air-power by the ratio of the two $(n_{rel}-1)$ factors. When the surrounding medium's index approaches the lens's own index, $(n_{rel}-1)\to 0$ and the lens's power collapses towards zero (it stops behaving like a lens at all when the indices match).

Step-by-Step Solution

1. In air: $P_{air} = (n_{lens}-1)\,G = (1.5-1)G = 0.5\,G = -5\ D \Rightarrow G = -10$ (in the appropriate units), where $G=\dfrac1{R_1}-\dfrac1{R_2}$.
2. In the liquid: relative index $n_{rel} = \dfrac{n_{lens}}{n_{liquid}} = \dfrac{1.5}{1.6}$, which is less than $1$ (liquid index exceeds lens index), so $(n_{rel}-1)$ is negative — a lens that was diverging in air can become converging in a denser medium. …