Step 1: Theoretical Analysis of Image Characteristics

To identify the optical device, we must analyze the given constraints:

• Image Nature: Virtual and erect.
• Image Size: Diminished (smaller than the object).
• Image Position: Between focus ($F$) and optical centre ($O$) on the same side as the object.

In optics, both spherical mirrors and lenses produce specific image types based on object placement. We evaluate the candidates:

Step 2: Validation via Lens Formula and Ray Properties

For a Concave Lens, the image is formed on the same side as the object [Per the Sign Convention for Spherical Lenses]. When an object is placed anywhere between infinity and the optical centre ($O$):

• A ray parallel to the principal axis appears to diverge from the principal focus ($F_1$).
• A ray passing through the optical centre ($O$) emerges undeviated.
• The intersection of these two rays always occurs between $F_1$ and $O$, resulting in a diminished, virtual, and erect image [Based on the properties of diverging lenses].

Step 3: Visual Representation

Step 4: Conclusion

Since the image is virtual, erect, diminished, and located on the same side of the device as the object for any position of the object, the optical device must be a Concave Lens (also known as a diverging lens).

Final Solution: The optical device is a Concave Lens.

Step 1: Identifying the Optical System

The window pane acts as a near object, while the distant building acts as an object at infinity ($u = \infty$). When the sharp image of the window pane is formed at $v = 15\text{ cm}$, the lens is focusing light from a finite distance. When the object distance increases from a finite value to infinity, the rays of light become parallel [Per the laws of refraction for convex lenses].

Step 2: Calculating Focal Length ($f$)

Since the initial image of the window is at $v = 15\text{ cm}$, and the window is at a finite distance, the focal length is not necessarily $15\text{ cm}$. However, for a distant object at infinity, the image is formed precisely at the principal focus ($F$).

The lens formula is given by:

As the object moves to infinity ($u \to \infty$):

Because the rays from a distant building are parallel, the new image distance $v_{new}$ will be equal to the focal length $f$. Since $15\text{ cm}$ was the image distance for a nearby object, the image for a distant object will form closer to the optical center than the original image position. Therefore, $f < 15\text{ cm}$.

Step 3: Determining the Direction of Screen Movement

According to the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$, as the object distance $u$ increases (moving from the window pane to the distant building), the image distance $v$ must decrease to maintain the equality for a constant focal length $f$.

Mathematically, if $u_1 < u_2$, then $v_1 > v_2$. Since $v_2 = f$ and $v_1 = 15\text{ cm}$, the screen must be moved towards the lens to capture the sharp image.

Step 4: Visualization of Focal Shift

Step 5: Conclusion

• Direction of Movement: The screen must be moved towards the lens.
• Approximate Focal Length: Since the lens focused a nearby object at $15\text{ cm}$, the focal length must be slightly less than $15\text{ cm}$.

Final Solution: Sudha must move the screen towards the lens; the approximate focal length of the lens is less than $15\text{ cm}$.

Step 1: Theoretical Foundation of Concave Lens Optics

A concave lens is a diverging lens. By definition, a ray of light parallel to the principal axis, after refraction through a concave lens, appears to diverge from the principal focus ($F_1$) on the same side of the lens [Per the Sign Convention for Lenses]. When an object is placed at the focus ($F_1$) of a concave lens, the ray passing through the optical center ($O$) undergoes no deviation [Per the Law of Optical Center passage].

Step 2: Ray Diagram Construction (Object at Focus)

Step 3: Analytical Breakdown of Image Characteristics

Based on the geometric construction, we apply the Thin Lens Formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.

• Object Distance ($u$): $-f$ (since the object is at the principal focus).
• Lens Equation Application:
  $\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$
  $\frac{1}{v} = \frac{1}{f} + \frac{1}{-f}$
  $\frac{1}{v} = 0 \implies v \to \infty$ [Mathematical limit definition].

Step 4: Conclusion of Image Properties

Final Solution: When an object is placed at the focus of a concave lens, the image is formed between the optical center ($O$) and the focus ($F_1$) on the same side of the lens. The image is virtual, erect, and highly diminished.

1. Given Parameters & Setup

An object is placed at a position $u > R$, where $R$ is the radius of curvature of a concave mirror. In spherical mirror geometry, this corresponds to the region between the Centre of Curvature ($C$) and infinity, but specifically "a little beyond $C$".

• Object Position: Between $C$ and $\infty$.
• Expected Image Characteristics:
  • Position: Between $C$ and $F$.
  • Nature: Real and Inverted.
  • Size: Diminished.

2. Geometric Ray Tracing Protocol

To determine the image location, we apply the rules of paraxial rays [Per Law of Reflection]:

1. Ray 1: A ray parallel to the principal axis, after reflection, passes through the Principal Focus ($F$).
2. Ray 2: A ray passing through the Principal Focus ($F$) becomes parallel to the principal axis after reflection.
3. Intersection: The point where these reflected rays meet defines the image position ($A'B'$).

3. Ray Diagram Construction

4. Mathematical Verification via Mirror Formula

Using the Cartesian sign convention [$ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} $]:

Since $u$ is beyond $C$ ($|u| > |2f|$), rearranging the formula $v = \frac{uf}{u-f}$ yields:

$f < v < 2f$

This mathematically confirms the image must form between the Principal Focus and the Centre of Curvature [Per Thin Mirror Approximation].

Final Solution: The image formed is Real, Inverted, and Diminished, located between the Centre of Curvature ($C$) and the Principal Focus ($F$) of the concave mirror.

Given Variables & Initial Setup

In a concave lens (a diverging lens), the image formation is dictated by the refractive properties of the lens surface. Given parameters:

• Lens Type: Concave (Diverging)
• Object Position ($u$): Between the principal focus ($F_1$) and twice the focal length ($2F_1$).
• Conventions: Using the Cartesian Sign Convention, light travels from left to right. The optical center ($O$) is the origin $(0,0)$.

Theoretical Principles of Ray Tracing

To locate the image, two primary rays are traced [Per the Rules for Image Formation by Concave Lenses]:

1. Ray 1: A ray parallel to the principal axis appears to diverge from the principal focus ($F_1$) after refraction. [Law of Refraction for Parallel Rays].
2. Ray 2: A ray passing through the optical center ($O$) of the lens proceeds without any deviation. [Property of Optical Center].

Visual Representation of Ray Diagram

Analytical Characteristics of the Image

Based on the intersection of the refracted ray and the virtual extension of the parallel ray, the image formed exhibits the following properties:

Mathematical Verification

Per the Gaussian Lens Formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.

Given the object is placed at $u$ such that $f < |u| < 2f$, and knowing that for a concave lens $f$ is negative ($-f$), we derive:

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

Since both $u$ and $f$ are negative, the resulting image distance $v$ must also be negative, confirming the image is formed on the same side of the lens as the object (virtual).

Final Solution: The image formed by a concave lens when the object is between $F_1$ and $2F_1$ is virtual, erect, diminished, and located between the optical center and the principal focus on the same side of the lens.

Step 1: Identifying Given Parameters and Sign Convention

Based on the problem statement, we identify the following variables according to the Cartesian Sign Convention:

• Magnification ($m$) = $-3$ (The image is real, as it is obtained on a screen, thus magnification is negative) [Per the definition of linear magnification for real images].
• Distance of the image from the lens ($v$) = $+80\text{ cm}$ (Since the image is real and formed on the opposite side of the lens, $v$ is positive).
• Distance of the object ($u$) = $?$
• Nature of the lens: Since the image is real and formed on a screen, the lens must be a Convex Lens.

Step 2: Calculating Object Distance ($u$)

The magnification ($m$) of a lens is defined as the ratio of image distance ($v$) to object distance ($u$):

$m = \frac{v}{u}$

Substituting the known values into the equation:

$-3 = \frac{80}{u}$

$u = \frac{80}{-3} = -26.67\text{ cm}$

[By the lens formula magnification derivation: $m = \frac{v}{u}$ for thin lenses]. The negative sign indicates that the object is placed in front of the lens.

Step 3: Geometrical Representation

Step 4: Determining Nature of the Image and Lens

We analyze the properties determined in the previous steps:

Final Solution: The candle should be placed at a distance of $26.67\text{ cm}$ from the lens. The lens is a convex lens, and the image is real and inverted.

Theoretical Foundation & Optical Principles

To produce a parallel beam of light from a point source, the incident light rays must be refracted or reflected in a manner that renders them parallel to the principal axis. This requires the point source to be positioned at the principal focus ($F$) of the optical device. When rays diverge from the focus and strike a parabolic or spherical mirror/lens, they emerge parallel [Per the laws of Geometrical Optics].

Step 1: Analyzing Optical Devices

• Concave Mirror: When a point source is placed at the principal focus $F$ of a concave mirror, the reflected rays travel parallel to the principal axis [By the Law of Reflection: $\angle i = \angle r$].
• Convex Lens: When a point source is placed at the principal focus $F$ of a convex lens, the refracted rays emerge parallel to the principal axis [Per Snell's Law and the principle of reversibility of light].

Step 2: Visual Representation of Parallel Beam Formation

Step 3: Comparative Analysis

Conclusion: Both a concave mirror and a convex lens can generate a parallel beam of light when a point source is placed at their respective principal foci. In the standard context of this curriculum, usually referring to specific configurations, both answers are valid depending on the optical setup provided.

Final Solution: Both a concave mirror and a convex lens produce a parallel beam of light when a point source is placed at their principal focus.

Step 1: Identifying the Geometric Conditions

The problem provides three critical constraints for the optical device:

• Nature of Image: Virtual and erect.
• Positional Constraint: The object is placed between the focus ($F$) and the optical center/pole of the device.
• Magnification Property: The image is enlarged and formed on the same side as the object.

Step 2: Analytical Evaluation of Potential Optical Devices

We evaluate the behavior of mirrors and lenses under the specified constraints:

Step 3: Theoretical Justification

For a Concave Mirror: Per the laws of reflection, when the object distance $u < f$, the light rays diverge after reflection. When produced backwards, they intersect behind the mirror, creating a virtual image where the magnification $m = -v/u$ is positive and greater than 1 [Mirror Formula: $1/f = 1/v + 1/u$].

For a Convex Lens: Per the refraction properties, when the object distance $u < f$, the rays emanating from the object diverge after passing through the lens. The virtual image is formed on the same side as the object [Lens Formula: $1/f = 1/v - 1/u$].

Step 4: Visualization

Final Solution: The optical device can be either a Concave Mirror or a Convex Lens.

Step 1: Theoretical Relationship between Power and Focal Length

The power ($P$) of a lens is defined as the reciprocal of its focal length ($f$) when expressed in meters. Mathematically, this is represented by the formula:

$P = \frac{1}{f (\text{in meters})}$

[Per the definition of optical power in the International System of Units (SI)]

The SI unit of power is the Dioptre ($D$), where $1D = 1m^{-1}$. This relationship implies that power is inversely proportional to the focal length: as the focal length increases, the power decreases, and as the focal length decreases, the power increases.

Step 2: Analysis of the Given Lenses

We are provided with two convex lenses (assuming convergence) with the following specifications:

• Lens A: $f_1 = 20\text{ cm} = 0.2\text{ m}$
• Lens B: $f_2 = 40\text{ cm} = 0.4\text{ m}$

Calculating the power for each lens:

$P_1 = \frac{1}{0.2\text{ m}} = +5.0\text{ D}$

$P_2 = \frac{1}{0.4\text{ m}} = +2.5\text{ D}$

Step 3: Visual Representation of Convergence

Step 4: Determining the Degree of Convergence

The convergence of a lens is directly proportional to its ability to bend incoming parallel rays of light. A lens with higher power has a greater ability to deviate light rays towards the principal axis [By the Law of Refraction/Snell's Law].

Since the power of Lens A ($5.0\text{ D}$) is greater than the power of Lens B ($2.5\text{ D}$), Lens A possesses a higher capacity to converge light rays.

Final Solution: The power of a lens is the reciprocal of its focal length ($P = 1/f$). To obtain more convergent light, you must use the lens with the shorter focal length, which is the 20 cm lens, as it has a higher optical power.

Step 1: Theoretical Analysis of Image Properties

To identify the optical device, we must evaluate the provided characteristics of the image:

• Nature of Image: Virtual and erect.
• Magnification: Enlarged ($|m| > 1$).
• Position of Object: Between the optical center/pole and the principal focus.
• Position of Image: Behind the device.

[Per the Sign Convention for Spherical Mirrors and Lenses]:

Step 2: Geometrical Verification via Ray Diagram

The configuration where an object placed between the focus and the optical center yields an enlarged, virtual image behind the device applies to both a Concave Mirror and a Convex Lens.

Step 3: Logical Conclusion

In the case of a Concave Mirror, when an object is placed between the Pole ($P$) and the Focus ($F$), the reflected rays appear to diverge from a point behind the mirror. This results in a virtual, erect, and magnified image [Per the Mirror Formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$].

Similarly, for a Convex Lens, when an object is placed between the Optical Center ($O$) and the Focus ($F_1$), the refracted rays diverge, creating a virtual, erect, and magnified image on the same side as the object [Per the Lens Formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$].

Final Solution: The device can be either a Concave Mirror or a Convex Lens.

Step 1: Theoretical Foundation & Ray Tracing Rules

To determine the image formation for an object placed at the centre of curvature ($C$) of a concave mirror, we apply the standard rules of spherical mirrors [Ref: Laws of Reflection and Geometry of Spherical Mirrors]:

• Rule 1: A ray parallel to the principal axis after reflection passes through the principal focus ($F$).
• Rule 2: A ray passing through the principal focus ($F$) after reflection travels parallel to the principal axis.

Step 2: Geometric Representation (Ray Diagram)

Step 3: Analytical Breakdown of Properties

When the object is placed at $C$, the reflection geometry yields the following analytical results:

Step 4: Algebraic Justification

Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

Given that the radius of curvature $R = 2f$. For an object at $C$, the object distance $u = -R = -2f$ [Per Sign Convention].

$\frac{1}{v} + \frac{1}{-2f} = \frac{1}{-f}$

$\frac{1}{v} = \frac{1}{2f} - \frac{1}{f} = \frac{1 - 2}{2f} = -\frac{1}{2f}$

$v = -2f$

Magnification ($m$) calculation: $m = -\frac{v}{u} = -\frac{-2f}{-2f} = -1$. The negative sign indicates an inverted image, and the magnitude of 1 indicates equality in size.

Final Solution: When the object is placed at the centre of curvature (C) of a concave mirror, a real, inverted image of the same size as the object is formed exactly at the centre of curvature (C).

1. Given Variables & Optical Configuration

An object is placed between the optical center ($O$) and the principal focus ($F_1$) of a convex lens. The parameters for this configuration are defined as follows:

• Lens Type: Convex (Converging lens)
• Object Position: $0 < u < f$ (where $u$ is the object distance and $f$ is the focal length).
• Expected Image Nature: Virtual, Erect, and Magnified.

2. Ray Diagram Construction

To locate the image, two specific light rays are utilized based on the standard rules of refraction for a convex lens:

1. Ray 1: A ray starting from the top of the object, traveling parallel to the principal axis. After refraction, it passes through the principal focus ($F_2$) on the other side of the lens.
2. Ray 2: A ray passing through the optical center ($O$) of the lens. This ray passes undeviated [Per the Property of Optical Center].

3. Analytical Derivation of Image Properties

By tracing the refracted rays backward, they appear to diverge from a point behind the object. This intersection point defines the position of the image.

Final Solution: The image formed is virtual, erect, and magnified, located on the same side of the lens as the object.

Step 1: Analyzing the Image Properties

The problem provides three critical descriptors for the image formed by the optical device:

• Nature of Image: Virtual and Erect.
• Size of Image: Diminished (smaller than the object).
• Position of Image: Between the pole ($P$) or optical center ($O$) and the focus ($F$), behind the device.

Per the rules of geometrical optics, we must evaluate the candidate devices:

Step 2: Evaluating the Geometric Constraints

The prompt specifies that the image is formed behind the device. By convention [Sign Convention for Spherical Mirrors and Lenses]:

• In a Convex Mirror, the image is formed behind the mirror (in the virtual space).
• In a Concave Lens, the image is formed on the same side as the object.

Since the problem explicitly states the image is formed "behind" the device, the device must be a convex mirror. In a convex mirror, for any object placed between infinity and the pole, the image is always formed between the pole ($P$) and the principal focus ($F$) behind the mirror, and it is always diminished and erect [Ray Optics Postulate].

Step 3: Geometrical Representation

Conclusion

Because the image is virtual, erect, diminished, and located behind the reflecting surface, the device must be a Convex Mirror.

Final Solution: The optical device is a Convex Mirror.

Theoretical Foundation: Image Formation at Infinity

An object placed at infinity implies that incident rays are parallel to the principal axis of the optical system [Per the Paraxial Approximation]. For both spherical mirrors and thin lenses, these parallel rays converge (or appear to diverge) at the principal focus ($F$).

Analysis of Optical Systems

Geometric Representation (Concave Mirror Case)

Logical Conclusion

Based on the ray optics principles defined by the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ and lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$, as $u \to \infty$, the term $\frac{1}{u} \to 0$. Thus, $v = f$ [Derivation from Gaussian optics].

In all standard spherical optical elements (Concave Mirror, Convex Mirror, Concave Lens, and Convex Lens), an object placed at infinity results in an image at the focal point. Consequently, the image size $h'$ is related to object size $h$ by magnification $m = \frac{h'}{h} = \frac{v}{u}$. Since $u \to \infty$, $m \to 0$, confirming that the image is highly diminished and point-sized in all four cases.

Final Solution: The image of an object placed at infinity is highly diminished and point-sized in all spherical mirrors (concave and convex) and all thin lenses (concave and convex).

1. Theoretical Foundation & Given Parameters

To determine the image formation of an object placed between the focus ($F$) and the centre of curvature ($C$) of a concave mirror, we apply the geometric laws of reflection. The image properties are governed by the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ and the magnification formula $m = \frac{h_i}{h_o} = -\frac{v}{u}$.

• Object Position: Between $F$ and $C$ ($f < |u| < 2f$).
• Image Characteristics: Real, inverted, and magnified.
• Image Position: Beyond the centre of curvature ($C$).

2. Ray Diagram Construction

The diagram below utilizes two primary light rays for construction:

1. Ray 1: Parallel to the principal axis; upon reflection, it passes through the principal focus ($F$).
2. Ray 2: Passing through the principal focus ($F$); upon reflection, it travels parallel to the principal axis [Per the Law of Reversibility of Light].

3. Analytical Justification of Image Properties

4. Step-by-Step Proof of Position

Using the mirror formula: $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$.

Given $f < |u| < 2f$, let us assume $f = -10\,\text{cm}$ and $u = -15\,\text{cm}$.

• $\frac{1}{v} = \frac{1}{-10} - \frac{1}{-15}$
• $\frac{1}{v} = \frac{-3 + 2}{30} = -\frac{1}{30}$
• $v = -30\,\text{cm}$

Since $|v| = 30\,\text{cm}$ and the centre of curvature is at $2f = 20\,\text{cm}$, the image distance $30 > 20$, confirming the image forms beyond $C$.

Final Solution: The image formed is Real, Inverted, and Magnified, located beyond the centre of curvature ($C$) of the concave mirror.

Theoretical Foundation: Refraction through Convex Lenses

To determine the correct ray diagram for a ray of light incident on a convex lens, we apply the geometric rules of ray optics [Per the Sign Convention and Lens Laws]:

• Rule 1: A ray of light parallel to the principal axis of a convex lens, after refraction, passes through the principal focus ($F_2$) on the other side of the lens.
• Rule 2: A ray of light passing through the optical center ($O$) of the lens emerges without any deviation.
• Rule 3: A ray of light passing through the principal focus ($F_1$) becomes parallel to the principal axis after refraction.

Step 1: Analyzing the Incident Geometry

In the provided scenario, the incident ray is depicted as being parallel to the principal axis of a convex lens. According to the standard geometric construction for image formation by a spherical lens:

If the incident ray $I$ is parallel to the principal axis, then the refracted ray $R$ must intersect the principal axis exactly at the principal focus $F_2$ located on the opposite side of the lens [By the Law of Refraction for Thin Lenses].

Step 2: Visual Representation

Step 3: Verification of the Correct Diagram

By evaluating the standard options provided in typical academic contexts for this question:

Final Solution: The correct ray diagram is the one where the incident ray, traveling parallel to the principal axis, passes through the principal focus ($F_2$) of the convex lens after refraction.