A plano-convex lens is made of material of refractive index n. When a small object is placed 30 cm a

Q: Why did we assume the image formed by refraction is real?

If the image were virtual, the magnification would be $+2$, leading to $v = -60\text{ cm}$. This would result in a focal length of $60\text{ cm}$, which would give a refractive index of $1.5$. Since $n=1.5$ is not among the options, the intended scenario is a real image.

Q: Why does the convex surface act as a convex mirror?

When light enters a medium, a small portion of it is always reflected at the boundary. The curved front surface of the lens reflects light just like a standalone convex mirror would.

Q: Why is the radius of the second surface infinity?

The problem states it is a plano-convex lens. A plane surface is essentially a sphere with an infinite radius of curvature, so $R_2 = \infty$.

Q: Why is the faint image virtual?

Convex mirrors always form virtual, erect, and diminished images for real objects placed in front of them. The faint image is formed by reflection from the convex surface, so it must be virtual.

Q: Can we use the Lens Maker's formula if the lens is thick?

The problem implies a 'thin' plano-convex lens by not providing a thickness. For thin lenses, the standard Lens Maker's formula is perfectly valid and accurate.

Question

A plano-convex lens is made of material of refractive index $n$. When a small object is placed $30	ext{ cm}$ away in front of the curved surface of the lens, an image of double the size of the object is produced. Due to reflection from the convex surface of the lens, another faint image is observed at a distance of $10	ext{ cm}$ away from the lens. Which of the following statement(s) is (are) true?

Options:

(A) The refractive index of the lens is $2.5$

(B) The radius of curvature of the convex surface is $45	ext{ cm}$

(C) The faint image is erect and real

(D) The focal length of the lens is $20	ext{ cm}$

Sigma Percentile · Accepted Answer

The journey of mastering optics often brings us to problems that beautifully intertwine multiple concepts. This problem from JEE Advanced 2016 is a perfect example, blending refraction through a lens with reflection from a curved surface. Let's dive into the fascinating physics behind it!

### Analyzing the Setup
Imagine a plano-convex lens with its curved surface facing an object placed $30	ext{ cm}$ away. The problem states that the image produced is double the size of the object. This gives us a crucial clue about the magnification.

Since the options suggest a focal length of $20	ext{ cm}$, and the object is at $30	ext{ cm}$ (which is between $f$ and $2f$), the image must be real and inverted. If we assumed a virtual image, the math would lead us to a refractive index not present in the options.

### The Refraction Phase
For a real, inverted image that is double the size, the magnification $m$ is $-2$. Using the magnification formula for lenses:
$$m = \frac{v}{u}$$
$$-2 = \frac{v}{-30}$$
$$v = +60	ext{ cm}$$

Now that we have the image distance, we can deploy the thin lens formula to find the focal length $f$:
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$
$$\frac{1}{60} - \frac{1}{-30} = \frac{1}{f}$$
$$\frac{1}{60} + \frac{2}{60} = \frac{1}{f}$$
$$\frac{3}{60} = \frac{1}{f} \implies f = +20	ext{ cm}$$
This immediately confirms that option (d) is correct!

### The Reflection Twist
Here is where the problem gets incredibly interesting. The curved front surface of the lens doesn't just refract light; it also reflects a small portion of it, acting exactly like a convex mirror!

This reflection forms a faint virtual image $10	ext{ cm}$ behind the mirror. For this convex mirror, the object distance $u$ is $-30	ext{ cm}$, and the image distance $v$ is $+10	ext{ cm}$. Let's use the mirror formula to find its focal length $f_m$:
$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f_m}$$
$$\frac{1}{10} + \frac{1}{-30} = \frac{1}{f_m}$$
$$\frac{3}{30} - \frac{1}{30} = \frac{1}{f_m}$$
$$f_m = +15	ext{ cm}$$

Since the radius of curvature $R$ is twice the focal length for a mirror, we have:
$$R = 2f_m = 30	ext{ cm}$$
This tells us that option (b) is incorrect. Also, since convex mirrors always form virtual and erect images for real objects, the faint image is virtual, making option (c) incorrect.

### The Master Equation
Finally, we need to find the refractive index $n$ of the lens material. We have the focal length of the lens ($f = 20	ext{ cm}$) and the radius of curvature of its first surface ($R_1 = +30	ext{ cm}$). Since it's a plano-convex lens, the second surface is flat, meaning $R_2 = \infty$.

We bring in the powerful Lens Maker's Formula:
$$\frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}ight)$$
$$\frac{1}{20} = (n - 1)\left(\frac{1}{30} - \frac{1}{\infty}ight)$$
$$\frac{1}{20} = (n - 1)\left(\frac{1}{30} - 0ight)$$
$$n - 1 = \frac{30}{20} = 1.5$$
$$n = 2.5$$

And there we have it! The refractive index of the lens is $2.5$, confirming that option (a) is also correct.

**Final Answer:** The correct statements are (a) and (d). This problem is a brilliant reminder to always look for the hidden phenomena—like reflection at a refracting surface—that make physics so deeply interconnected and beautiful.

Select Answer:

Visualized Solution

$S e t u p$

$M a g ni f i c a t i o n$

$L e n s F or m u l a$

$F oc a l L e n g t h$

$R e f l ec t i o n f r o m C u r v e d S u r f a ce$

$M i r r or F or m u l a$

$R a d i u s o f C u r v a t u r e$

$L e n s M ak e r^{'} s F or m u l a$

$R e f r a c t i v e I n d e x$

$C o n c l u s i o n$

The Sigma Insight: Refraction at Spherical Surface

Analyzing the Setup Imagine a plano-convex lens with its curved surface facing an object placed $30 cm$ away

The Refraction Phase For a real, inverted image that is double the size, the magnification $m$ is $- 2$

The Reflection Twist Here is where the problem gets incredibly interesting

The Master Equation Finally, we need to find the refractive index $n$ of the lens material

Similar Questions

In the figure, light is incident on a thin lens as shown. The radius of curvature for both the surfaces is $R$ . Determine the focal length of this system.

A point object is placed at a distance of 20 cm from a thin plano-convex lens of focal length 15 cm. The plane surface of the lens is now silvered. The image created by the system is at

A point object is placed at the centre of a glass sphere of radius $6 cm$ and refractive index $1.5$ . The distance of the virtual image from the surface of the sphere is

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnification

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Lens Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Focal Length

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Reflection from Curved Surface

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mirror Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Radius of Curvature

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Lens Maker′s Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Refractive Index

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion