Lenses (converging and diverging) | ICSE Class 10 Physics Notes

This note covers converging and diverging lenses, their action as prisms, optical terms, construction rays, image formation, magnifying glasses, applications, sign conventions, the lens formula and lens power.

What are converging and diverging lenses?

A lens is a transparent material bounded by two surfaces, at least one of which is spherical, meaning that it forms part of a sphere. A transparent material transmits light. A ray is a line representing the direction in which light travels.

Refraction is the change in direction that light can undergo when it passes between transparent media, such as air and glass. A medium is a substance through which light travels. Lenses form images by refraction at their surfaces.

How do their shapes differ?

A double convex lens has two surfaces bulging outwards. It is thicker at the middle than at its edges. A double concave lens has two surfaces curved inwards and is thinner at the middle than at its edges.

For the glass lenses in air considered here, a convex lens brings parallel rays together and is called a converging lens. A concave lens spreads parallel rays apart and is called a diverging lens. Convergence means coming together; divergence means spreading apart.

How does a set of prisms explain a lens?

A prism is a transparent optical element with inclined plane refracting faces. In its usual triangular cross-section, the base is opposite the vertex where the refracting faces meet. A glass prism in air deviates transmitted light towards its base; deviation means a change from the original direction.

Imagine a convex lens divided into narrow prism-like portions. Their bases face towards the middle, so rays above and below the central line bend towards one another. For a concave lens, the bases face towards the edges, so these rays spread apart.

This is a way to understand the lens's action, not a statement that a lens is assembled from separate prisms. Its smoothly curved surfaces provide the changing inclinations that the prism model represents.

What do the principal axis, foci and focal plane mean?

Each spherical surface of a lens belongs to an imaginary sphere. The centres of these spheres are the centres of curvature, labelled C₁ and C₂. Their radii are the radii of curvature, labelled R₁ and R₂ respectively.

The principal axis is the straight line through the centres of curvature. The optical centre, labelled O, is the central point of the lens. For a thin lens, whose thickness is neglected in the ray construction, a ray passing through O emerges without deviation.

The aperture is the effective diameter of the lens's circular outline. We consider thin lenses with small apertures, for which this diameter is much less than the radius of curvature. The diagrams represent refraction by such lenses in a simplified way.

Where are the two principal foci?

For a convex lens, rays travelling parallel to the principal axis meet after refraction at a point on that axis. This point is a principal focus. For a concave lens, the refracted rays appear to diverge from a principal focus on the incident side.

A lens has two principal foci because light can enter from either side. Here F₁ denotes the focus on the left and F₂ the focus on the right. Unless otherwise specified, incident light, meaning light approaching the lens, travels from left to right.

Definition: The focal length, represented by f, is the distance from the optical centre to a principal focus. The SI unit of focal length is the metre, written m; SI denotes the International System of Units.

The focal plane is the plane perpendicular to the principal axis through a principal focus. Parallel rays inclined to the axis meet, or appear to come from, a point in this plane. The principal focus is its intersection with the axis.

What do the diagram labels mean?

The labels 2F₁ and 2F₂ mark positions at twice the focal distance from O on the left and right respectively. They are useful reference positions for image formation. They should not be treated as definitions of the centres of curvature.

The upper drawing shows parallel rays meeting to the right of a convex lens at F₂. The lower drawing shows rays spreading after a concave lens, with dotted backward extensions meeting at F₁ on the left.

See Fig. 9.12 in your NCERT textbook

How do construction rays locate an image?

Construction rays are rays whose refracted directions are known, allowing an image to be located. A small linear object is represented by an arrow perpendicular to the principal axis, with its base on the axis. Its tip supplies the rays used in the construction.

The incident ray approaches the lens; the refracted ray leaves it after transmission. Two rays from the same object point are sufficient to locate the corresponding image point. Using rays from different points of the object would not locate one image point.

Which three rays are useful?

In the convex-lens drawing, the incident ray passes through the left focus and emerges horizontally. In the concave-lens drawing, its dotted continuation points towards the right focus, while the emergent ray is parallel to the axis.

See Fig. 9.14 in your NCERT textbook

How should the construction be completed?

1. Draw the principal axis, lens and optical centre. Mark both foci and the positions at twice the focal distance.
2. Place the object arrow at the required position, with its base on the principal axis.
3. From the tip, draw a ray parallel to the axis and a second ray through O. Apply the appropriate lens rules.
4. Find where the refracted rays meet. If they spread apart, extend them backwards using dotted lines to locate their apparent intersection.
5. Draw the image arrow from the axis to this intersection. Compare its position, direction and height with the object.

A real image forms where refracted rays actually meet and can be obtained on a screen. A virtual image forms where their backward extensions meet and cannot be obtained on a screen at that apparent position.

Erect means the image has the same vertical orientation as the object; inverted means it is upside down relative to it. Diminished means smaller, and enlarged means larger. These descriptions concern different image properties and should be stated separately.

How does a convex lens form images at different object positions?

A convex lens can give different image positions, sizes and orientations as the object moves. The key reference positions are F₁ and 2F₁ on the object side. Read each case by tracing the rays, then describe the image's location, relative size and nature.

What are the standard object positions?

At infinity describes the limiting case of a very distant object whose rays reaching the lens are effectively parallel. The rays then converge at the focus. For a nearby object, rays from its tip reach the lens along different directions.

The drawings show upright object arrows on the left and inverted image arrows on the right. The image is smaller, equal-sized or larger as the object moves from beyond 2F₁ to between F₁ and 2F₁.

Reference: NCERT Class 10 Figure 9.16, panels b, c and d

What changes at and inside the focus?

With the object at F₁, rays from an object point emerge parallel. They do not meet at a finite distance, so a sharp image is not obtained on a screen at a finite position. This is the limiting situation described as an image at infinity.

Move the object between F₁ and O, and the emerging rays spread apart. Their backward extensions meet on the object side, giving an erect, enlarged, virtual image. Looking through the lens allows this image to be seen, although a screen cannot capture it there.

Note: Converging action does not mean every object produces a real image. The image also depends on object position. A convex lens produces a virtual image when the object is within its focal length.

When checking a diagram, relate the height of the image arrow to its distance from the axis. An arrow below the axis represents an inverted image of an upright object. A longer arrow represents enlargement, regardless of which side of the lens it occupies.

How does a concave lens form an image?

For a real object placed in front of a concave lens, the refracted rays spread apart. A real object here means an object from which light reaches the lens, as in the candle arrangement. The image is located using backward extensions of the refracted rays.

The concave lens gives a virtual, erect and diminished image for these object positions. Its image lies on the object side of the lens. Unlike the convex-lens cases, moving this object does not produce an inverted image on the other side.

Where is the image located?

The first drawing shows parallel rays diverging, with dotted extensions meeting at F₁. The second shows an upright object and a smaller upright image between F₁ and O. Dotted extensions locate the image on the left of the lens.

See Fig. 9.17 in your NCERT textbook

How is the virtual image constructed?

Draw a ray from the object tip parallel to the principal axis. After refraction, draw it away from the axis so that its backward extension passes through F₁. Draw the second ray through O without deviation.

The backward extension of the first refracted ray meets the line of the undeviated ray on the object side. That intersection locates the image tip. The image base remains on the principal axis, directly below the tip for an upright image.

The word appears matters: the refracted light does not actually travel backwards through the virtual image. Solid lines with arrows represent light paths; dotted backward extensions are construction lines. Confusing the two would incorrectly suggest that the lens has formed a real image.

How does a magnifying glass work, and where are lenses used?

A magnifying glass, or simple microscope, is a converging lens of small focal length. Place the object between the lens and its principal focus, and view it through the lens from the opposite side. The image is erect, virtual and enlarged.

Magnification describes image size relative to object size. In a ray diagram, compare the image arrow with the object arrow. The larger upright image in the magnifying-glass construction shows enlargement without requiring the image to be projected onto a screen.

How does the ray diagram explain enlargement?

1. Draw a convex lens and mark its optical centre and both principal foci.
2. Place a small upright object between F₁ and O, with its base on the axis.
3. Draw a parallel ray from its tip, refracted through F₂, and a ray through O without deviation.
4. Extend the emerging rays backwards until they intersect on the object side.
5. Draw the upright image arrow at this intersection. It is taller than the object arrow, showing enlargement.

The small upright object is between the left focus and O. The refracted rays spread apart on the right. Dotted extensions meet at the top of a taller upright image on the left.

Reference: NCERT Class 10 Figure 9.16, panel f

If the object is slightly less than one focal length away, its image is virtual and closer than infinity. With the object at the focus, the emerging rays are parallel. An image at infinity is often considered most suitable for viewing by the relaxed eye.

What applications depend on lens action?

A watchmaker uses a magnifying glass to see tiny parts. Lenses are also used in spectacles and in optical instruments such as cameras, microscopes and telescopes. Instrument lens systems can combine lenses to improve magnification and image sharpness.

Myopia, or near-sightedness, means nearby objects are clear but distant objects are not distinct. A concave corrective lens of suitable power helps bring the image of a distant object onto the retina, the light-sensitive membrane inside the eye.

Hypermetropia, or far-sightedness, means distant objects are clear but nearby objects are not distinct. A convex lens of appropriate power provides additional convergence so that light from a nearby object forms its image on the retina. The required lens therefore depends on the defect.

How are signs assigned to lens distances?

The New Cartesian sign convention is a consistent rule for assigning positive and negative signs to distances and heights. For lenses, use the optical centre as the origin, meaning the point from which distances are measured, and the principal axis as the reference line.

Let u represent object distance and v represent image distance, both measured from O along the axis. The previously defined f represents focal length. In the usual drawing, the object is on the left and incident light travels towards the right.

The SI unit of object distance is the metre (m). The SI unit of image distance is also the metre (m). Both are lengths, like focal length; centimetres may be used instead when all three distances in the lens formula use that same unit.

Which directions are positive?

• Distances measured to the right of O, in the direction of incident light, are positive.
• Distances measured to the left of O, against the direction of incident light, are negative.
• Heights measured above the principal axis are positive; heights below it are negative.
• The focal length of a convex lens is positive; that of a concave lens is negative.

How should a calculated sign be interpreted?

A negative distance does not mean a physically negative separation. It specifies direction from the chosen origin. State both the result with its sign and its physical meaning, such as an image on the same side of the lens as the object.

Note: Assign signs before substitution. Do not turn the final answer positive merely because distances in everyday language are given as positive separations. Retain the algebraic sign until the image position has been interpreted.

How is the lens formula used in direct numerical problems?

The lens formula relates object distance u, image distance v and focal length f. All are measured from the optical centre, with signs assigned by the New Cartesian convention. Use the same length unit for all three quantities in one substitution.

Definition: For a thin spherical lens, the lens formula is $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$. Its rearrangements are $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}$ and $\frac{1}{u}=\frac{1}{v}-\frac{1}{f}$.

What sequence keeps a calculation clear?

1. Identify the lens type and assign the sign of its focal length.
2. Write each known distance with its sign and unit, using the stated side of the lens.
3. Select the rearrangement containing the required unknown on the left.
4. Substitute the values and combine the signed fractions before taking the reciprocal.
5. Interpret the answer using the ray-diagram case: identify the side and nature of the image.

A reciprocal is one divided by a quantity. Thus, an intermediate result for 1/v is not yet the answer for v. A distance and its reciprocal have different units, so recording the algebra helps prevent this mistake. Reciprocal centimetres are written cm⁻¹, where cm denotes centimetre.

Worked example 1. A concave lens has focal length 15 cm. Its virtual image is 10 cm from the lens on the object side. Find the object distance.

Given: f = −15 cm and v = −10 cm. Formula: 1/u = 1/v − 1/f.

Substitute: 1/u = −1/10 + 1/15 = −1/30 cm⁻¹.

Answer: u = −30 cm. The object is 30 cm in front of the lens. The image distance lies between O and the focus on the object side, as expected for a concave lens.

Worked example 2. An object is 15 cm in front of a convex lens of focal length 10 cm. Find the image distance and state its nature.

Given: u = −15 cm and f = +10 cm. Formula: 1/v = 1/f + 1/u.

Substitute: 1/v = 1/10 − 1/15 = 1/30 cm⁻¹.

Answer: v = +30 cm. The image is 30 cm from the lens on the opposite side. It is real and inverted. The object is between F₁ and 2F₁, matching an enlarged image beyond 2F₂.

In these calculations, cm⁻¹ means reciprocal centimetres. A result expressed as a reciprocal must be inverted to obtain a length in centimetres. Keeping this distinction visible is particularly useful when negative fractions are involved.

Derivation: How does refraction at two surfaces give the thin lens formula?

For this supplementary derivation, consider a thin lens with small aperture and rays close to the principal axis. Let $n_1$ be the refractive index of the surrounding medium on both sides and $n_2$ that of the lens. Use signed radii $R_1$ and $R_2$.

1. At the first surface, light passes from the surrounding medium into the lens. If $v_1$ is the intermediate image distance, refraction at a spherical surface gives $\frac{n_2}{v_1}-\frac{n_1}{u}=\frac{n_2-n_1}{R_1}$.
2. The intermediate image acts as the object for the second surface. Neglecting lens thickness lets its distance from either surface be treated as the same. Refraction back into the surrounding medium gives $\frac{n_1}{v}-\frac{n_2}{v_1}=\frac{n_1-n_2}{R_2}$.
3. Add these equations. The intermediate image terms cancel, leaving $n_1\left(\frac{1}{v}-\frac{1}{u}\right)=(n_2-n_1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$.
4. For an object at infinity, the reciprocal object distance tends to zero and the image lies at the focus. Therefore $\frac{n_1}{f}=(n_2-n_1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$.
5. The right sides of the last two equations are identical. Equating their left sides and dividing by $n_1$ gives the relation between object distance, image distance and focal length.

Result: $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$. Apply the New Cartesian signs throughout. The thin lens formula holds for converging and diverging lenses, with real or virtual images, within the thin lens approximation.

What is lens power, and how is it calculated?

Power of a lens, represented by P, measures its ability to converge or diverge light and is the reciprocal of its focal length. A shorter focal length means stronger convergence for a convex lens or stronger divergence for a concave lens.

P = 1/f, with f in metres. The SI unit of power of a lens is the dioptre, written D. One dioptre is the power of a lens whose focal length is one metre: 1 D = 1 m⁻¹, where m⁻¹ means reciprocal metres.

The power of a convex lens is positive, and that of a concave lens is negative. The sign identifies the lens's converging or diverging action. When comparing strength, compare the magnitude, meaning the numerical size without the sign.

How are units handled?

Convert a focal length from centimetres to metres before finding power in dioptres: 1 m = 100 cm. To find focal length from power, use f = 1/P; power in dioptres then gives focal length in metres.

Note: A centimetre value cannot be substituted directly into P = 1/f to obtain dioptres.

Worked example 3. A convex lens forms a real, inverted image of a needle 50 cm from the lens. The image equals the needle in size. Find the needle position, focal length and power.

Given: The equal-sized real-image case places the object at 2F₁ and image at 2F₂. Therefore u = −50 cm and v = +50 cm.

Formula: 1/f = 1/v − 1/u; P = 1/f, using metres for power.

Substitute: 1/f = 1/50 − 1/(−50) = 1/25 cm⁻¹. Hence f = +25 cm = +0.25 m, and P = 1/0.25.

Answer: The needle is 50 cm in front of the lens; f = +25 cm and P = +4 D.

Worked example 4. A corrective lens has power +2.0 D. Find its focal length in metres and centimetres, and identify its type.

Formula: f = 1/P; f in centimetres = 100 × f in metres.

Substitute: f = 1/(+2.0) = +0.50 m. Converting the length gives 100 × 0.50 = 50 cm.

Answer: f = +0.50 m = +50 cm. The positive power identifies a convex, converging lens.

Worked example 5. A lens has power −2.5 D. Find its focal length and identify its type.

Formula: f = 1/P. Substitute: f = 1/(−2.5).

Answer: f = −0.40 m. Its negative focal length and power identify a concave, diverging lens.

Worked example 6. Find the power of a concave lens of focal length 2 m.

Given: f = −2 m. Formula: P = 1/f. Substitute: P = 1/(−2).

Answer: P = −0.5 D, corresponding to focal length −2 m. The negative sign represents the diverging action of the lens.

Use the lens type as a final check on the sign. A concave lens cannot have positive power in this treatment. A calculation producing the wrong sign should be checked for an incorrect focal-length sign or a lost negative during division.

Derivation: How do the powers of two thin lenses in contact combine?

For this supplementary derivation, take two thin lenses of focal lengths $f_1$ and $f_2$ in contact. Treat their optical centres as coincident. Let $v_1$ locate the image formed by the first lens alone and $v$ locate the final image.

1. Apply the lens formula to the first lens: $\frac{1}{v_1}-\frac{1}{u}=\frac{1}{f_1}$.
2. The first image acts as the object for the second lens. Since the lenses are in contact, its signed object distance is $v_1$. Thus $\frac{1}{v}-\frac{1}{v_1}=\frac{1}{f_2}$.
3. Add the equations to eliminate the intermediate image distance: $\frac{1}{v}-\frac{1}{u}=\frac{1}{f_1}+\frac{1}{f_2}$.
4. An equivalent single lens of focal length $f$ produces the same final image from the same object. Comparing with its lens formula gives $\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$. Express each focal length in metres to replace its reciprocal with power in dioptres.

Result: $P=P_1+P_2$. Add powers algebraically, retaining positive signs for converging lenses and negative signs for diverging lenses. This relation assumes thin lenses in contact.

Glossary

• Lens — A transparent optical material bounded by two surfaces, at least one of which is spherical.
• Convex lens — A lens thicker at the middle than the edges, converging parallel rays in air.
• Concave lens — A lens thinner at the middle than the edges, diverging parallel rays in air.
• Principal axis — The straight line passing through the two centres of curvature of a spherical lens.
• Optical centre — The central point of a thin lens through which a ray emerges without deviation.
• Centre of curvature — The centre of the imaginary sphere of which a lens surface forms a part.
• Principal focus — The axial point where parallel incident rays meet, or appear to originate, after refraction.
• Focal length — The distance between the optical centre of a lens and a principal focus.
• Focal plane — A plane perpendicular to the principal axis and passing through a principal focus.
• Real image — An image formed by actual meeting of refracted rays and obtainable on a screen.
• Virtual image — An image located by backward extensions of refracted rays, not obtainable on a screen there.
• Magnification — A comparison of image size with object size, showing enlargement, equality or diminution.
• Lens power — The reciprocal of focal length in metres, expressing convergence or divergence in dioptres.
• Dioptre — The unit of lens power, equal to the power of a lens of focal length one metre.

Common errors and misconceptions

• Misconception: A convex lens produces a real image for every object position. Correct: An object between its focus and optical centre produces an erect, enlarged, virtual image on the object side.
• Misconception: A concave lens's refracted rays actually pass through its virtual image. Correct: Their backward extensions intersect there. The refracted rays themselves spread apart after leaving the lens.
• Misconception: The focus and optical centre are interchangeable labels. Correct: The optical centre is the central reference point; a principal focus is located at the focal distance from it.
• Misconception: Image distances must be positive because distances cannot be negative. Correct: A negative sign specifies direction from the optical centre. A virtual image on the left has negative image distance.
• Misconception: A focal length in centimetres gives power in dioptres directly through its reciprocal. Correct: Convert the focal length to metres before calculating power in dioptres.
• Misconception: A magnifying glass gives an inverted image on a screen. Correct: With the object inside its focal length, it gives an erect, enlarged, virtual image viewed through the lens.
• Misconception: The lens formula can use unsigned object distance alongside signed focal length. Correct: Assign signs consistently to every distance before substitution, then interpret the resulting sign physically.

Exam-style questions with model answers

Q1. Distinguish a double convex lens from a double concave lens by thickness and action on parallel light in air. [2 marks]

1. A double convex lens is thicker at the middle than at its edges and converges parallel incident rays.
2. A double concave lens is thinner at the middle than at its edges and diverges parallel incident rays.

Q2. Explain how to locate the image of an upright object placed beyond twice the focal length of a convex lens. State its characteristics. [4 marks]

1. From the object tip, draw a ray parallel to the principal axis. After refraction, it passes through the focus on the opposite side.
2. Draw a second ray from the same tip through the optical centre; it emerges without deviation.
3. The refracted rays meet between the focus and twice the focal distance on the other side, locating the image tip.
4. The image is real, inverted and diminished, so it can be received on a screen at that position.

Q3. An object is 15 cm in front of a convex lens of focal length 10 cm. Light travels from left to right. Calculate the image distance and state its nature. [3 marks]

1. The object lies to the left, so u = −15 cm. The convex lens has positive focal length, f = +10 cm.
2. Using 1/v = 1/f + 1/u gives 1/v = 1/10 − 1/15 = 1/30 cm⁻¹, so v = +30 cm.
3. The positive result places the image 30 cm on the opposite side of the lens. It is real and inverted.

Q4. A concave lens has focal length 15 cm and forms a virtual image 10 cm from the lens on the object side. Light travels from left to right. Find the object distance. [3 marks]

1. For the concave lens, f = −15 cm. The virtual image is on the left, so its image distance is v = −10 cm.
2. Rearrange the lens formula as 1/u = 1/v − 1/f. Substitution gives 1/u = −1/10 + 1/15 = −1/30 cm⁻¹.
3. Taking the reciprocal gives u = −30 cm. The object must therefore be placed 30 cm in front of the lens.

Q5. Explain the arrangement and ray construction for a convex lens used as a magnifying glass, including the image's position, nature and relative size. [5 marks]

1. Choose a convex lens of small focal length. Place the small upright object between the principal focus on the object side and the optical centre.
2. Draw a ray from the object tip parallel to the principal axis. After refraction through the lens, it travels through the focus on the opposite side.
3. Draw another ray from the same tip through the optical centre. This ray emerges without deviation, providing the second construction ray.
4. The emerging rays spread apart. Extend them backwards until they meet on the object side of the lens; this intersection locates the image tip.
5. The image is virtual, erect and enlarged. View it through the lens from the opposite side; it cannot be obtained on a screen at its apparent position.

Q6. A convex lens forms a real, inverted image of a needle 50 cm from the lens. The image is the same size as the needle. Take incident light from left to right. Find the object position, focal length and power. [5 marks]

1. The equal-sized real image corresponds to the object at twice the focal distance on the incident side and the image at twice the focal distance on the other side.
2. Therefore the needle is 50 cm in front of the lens. With light travelling left to right, u = −50 cm and v = +50 cm.
3. Apply the lens formula: 1/f = 1/v − 1/u = 1/50 − 1/(−50) = 1/25 cm⁻¹.
4. Taking the reciprocal gives f = +25 cm. Convert this to metres before calculating power: f = +0.25 m.
5. Power P = 1/f = 1/0.25 = +4 D. The positive sign agrees with the converging action of the convex lens.

Q7. A concave lens has focal length 2 m. Calculate its power and explain the sign. [2 marks]

1. For the concave lens, f = −2 m, so P = 1/f = −0.5 D.
2. The negative power indicates that the lens diverges parallel incident rays.

Q8. For a real object at a finite distance in front of a concave lens, explain how two construction rays locate its image and state the image characteristics. [4 marks]

1. Draw a ray from the object tip parallel to the principal axis. It emerges as if from the focus on the object side.
2. Draw another ray from the same tip through the optical centre, without deviation.
3. The backward extension of the first refracted ray intersects the line of the second between the optical centre and the object-side focus.
4. This locates a virtual, erect and diminished image on the object side. The image cannot be obtained on a screen.

Key takeaways

• A convex lens converges parallel light in air, whereas a concave lens diverges it.
• The optical centre is the reference point for lens distances; focal length measures its separation from a principal focus.
• Two suitable rays from the same object point locate its image through actual intersection or backward extension.
• A convex lens produces different images depending on object position, including an enlarged virtual image when the object is inside its focal length.
• A concave lens produces a virtual, erect and diminished image of a real object in front of it.
• A magnifying glass uses a short-focal-length convex lens with the object inside its focal length for an enlarged virtual image.
• Use signed distances consistently in the lens formula, and interpret the final sign as a direction from the optical centre.
• Lens power is the reciprocal of focal length in metres, positive for convex lenses and negative for concave lenses.

Test yourself

Where does a ray through the optical centre of a thin lens go?

It emerges without deviation, continuing along the same straight line through the optical centre.

Where is the image when an object is at twice the focal distance of a convex lens?

It is at twice the focal distance on the opposite side, real, inverted and equal in size to the object.

What happens to rays from an object point at the principal focus of a convex lens?

They emerge parallel, so they do not form an image at a finite distance. This is the image-at-infinity limiting case.

Where is the image of a finite-distance real object in a concave lens?

It lies between the optical centre and the focus on the object side, and is virtual, erect and diminished.

Why does a magnifying-glass image fail to appear on a screen?

It is virtual: backward extensions intersect at the image position, but the refracted rays do not actually meet there.

A lens has power −2.5 D. What are its focal length and type?

Its focal length is 1/(−2.5) = −0.40 m. It is a concave, diverging lens.

What is the difference between a principal focus and a focal plane?

A principal focus is a point on the axis. A focal plane passes through that point perpendicular to the axis.

Which lens corrects myopia, and which corrects hypermetropia?

A concave lens of suitable power corrects myopia; a convex lens of appropriate power corrects hypermetropia.