A point object in air is in front of the curved surface of a plano-convex lens. The radius of curvat

Q: Why is the radius of the plane surface taken as infinity?

A plane surface can be thought of as a part of a sphere with an infinitely large radius. Hence, its curvature is zero, and its radius is infinity.

Q: Does it matter which side of the lens the light hits first?

No, the focal length of a thin lens is the same regardless of which side the light enters. The signs of $R_1$ and $R_2$ will swap, but the final result remains unchanged.

Q: What if the lens was placed in water instead of air?

The focal length would change! The Lens Maker's Formula uses the relative refractive index of the lens with respect to the surrounding medium. In water, the relative refractive index would be smaller, resulting in a longer focal length.

Q: Why is the focal length positive?

A positive focal length indicates that the lens is converging. Plano-convex lenses are thicker in the middle than at the edges, which makes them converging lenses.

Q: What is the significance of the term $(\mu - 1)$ in the formula?

This term represents the optical density of the lens relative to air. A higher refractive index means the lens bends light more strongly, leading to a shorter focal length.

Question

A point object in air is in front of the curved surface of a plano-convex lens. The radius of curvature of the curved surface is $30	ext{ cm}$ and the refractive index of the lens material is $1.5$, then the focal length of the lens (in cm) is ……… .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a point object placed in front of a plano-convex lens. The light rays first strike the curved surface, and then exit through the flat plane surface.

To find the focal length of this lens, we need the **Lens Maker's Formula**. This powerful equation connects the focal length to the refractive index of the material and the radii of curvature of its two surfaces.

### The Master Equation

The Lens Maker's Formula is given by:

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

Let's identify our given values. The refractive index $\mu$ is $1.5$. The first surface is curved with a radius of $30	ext{ cm}$. Since it bulges towards the object, its center of curvature lies to the right, making $R_1$ positive $30	ext{ cm}$.

The second surface is flat. A flat plane can be thought of as a sphere with an infinitely large radius, so its radius $R_2$ is $\infty$.

### Final Calculation

Now, let's substitute these values into our formula:

$\frac{1}{f} = (1.5 - 1) \left( \frac{1}{30} - \frac{1}{\infty} ight)$

We know that $\frac{1}{\infty}$ is $0$. And $1.5 - 1$ gives us $0.5$, or $\frac{1}{2}$. So the equation simplifies to:

$\frac{1}{f} = 0.5 	imes \frac{1}{30}$

$\frac{1}{f} = \frac{1}{2} 	imes \frac{1}{30}$

Multiplying the denominators, we get:

$\frac{1}{f} = \frac{1}{60}$

Therefore, the focal length of the lens is exactly **$60	ext{ cm}$**.

A positive focal length confirms it acts as a converging lens. Interestingly, even if the light entered from the flat side instead, swapping $R_1$ and $R_2$ would just flip the signs, but the focal length would remain exactly the same. The lens behaves identically from both sides!

Animated Solution for Physics - Optics: A point object in air is in front of the curved surface of a plano-convex lens. The radius of curvature of the curved surface is $30 cm$ and the refractive index of the lens material is $1.5$ , then the focal length of the lens (in cm) is ……… .

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

The Lens Maker's Formula

Identifying the Parameters

Substituting the Values

Simplifying the Equation

Final Calculation

The Way Forward

The Sigma Insight: Lens

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

A plano-convex lens of refractive index 1.5 and radius of curvature 30 cm is silvered at the curved surface. Now, this lens has been used to form the image of an object. At what distance from this lens, an object be placed in order to have a real image of the size of the object

Diameter of a plano-convex lens is $6 cm$ and thickness at the centre is $3 mm$ . If speed of light in material of lens is $2 \times 1 0^{8} m/s$ , the focal length of the lens is

A thin lens of refractive index 1.5 has a focal length of 15 cm in air. When the lens is placed in a medium of refractive index 4/3, its focal length will become ...... cm.

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Curved surfaces of a plano-convex lens of refractive index $μ_{1}$ and a plano-concave lens of refractive index $μ_{2}$ have equal radius of curvature as shown in figure. Find the ratio of radius of curvature to the focal length of the combined lenses.

The refractive index of a converging lens is $1.4$ . What will be the focal length of this lens if it is placed in a medium of same refractive index ? (Assume the radii of curvature of the faces of lens are $R_{1}$ and $R_{2}$ respectively)

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

The Lens Maker's Formula

Identifying the Parameters

Substituting the Values

Simplifying the Equation

Final Calculation

The Way Forward

The Sigma Insight: Lens

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

A plano-convex lens of refractive index 1.5 and radius of curvature 30 cm is silvered at the curved surface. Now, this lens has been used to form the image of an object. At what distance from this lens, an object be placed in order to have a real image of the size of the object

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Diameter of a plano-convex lens is 6 cm and thickness at the centre is 3 mm. If speed of light in material of lens is 2×108 m/s, the focal length of the lens is

A thin lens of refractive index 1.5 has a focal length of 15 cm in air. When the lens is placed in a medium of refractive index 4/3, its focal length will become ...... cm.

The same size images are formed by a convex lens when the object is placed at 20 cm or at 10 cm from the lens. The focal length of convex lens is ………… cm.

Diameter of a plano-convex lens is $6 cm$ and thickness at the centre is $3 mm$ . If speed of light in material of lens is $2 \times 1 0^{8} m/s$ , the focal length of the lens is