Solution: (i) When an object (AB) is located at infinity, parallel rays from the object appear to fall on the concave lens. It causes an image to form at focus. The scale of this image is greatly...

Solution: (i) When the object is situated at infinity, the position of the image is at F2. It is real, inverted and very much diminished in size. (ii) When the object (AB) is situated beyond 2F1,...

Type of lensPosition of objectNature of imageSize of imageConvexBetween the optical centre and focusConvexAt focusConcaveAt infinityConcaveAt any distance Solution: Type of lensPosition of...

(c)inverted and enlarged, (d)upright and enlarged? Solution: (c) Between F and 2F Object must be placed between F and 2F of a convex lens so that the image formed is inverted and enlarged. (d)...

(a)at infinity, (b)of same size as the object, Solution: (a) At focus, Object must be placed at the focus of a convex lens so that the image is formed at infinity. (b) At 2F, Object must be placed...

Solution: The virtual image formed by a convex lens will always be i) erect and ii) magnified.

Solution: The image formed by a concave lens is virtual and diminished.

Solution: No, a concave lens cannot produce a picture two times the size of the item because it diverges the incident rays, resulting in a decreased image.

(b)State whether the image in part (a) is magnified or diminished? Solution: (a) Since lens here always forms an erect and virtual image, the lens used here is a concave lens (b) The image formed by...

(a)Name the lens. (b)Draw a labelled ray diagram to show the image formation. Solution: (a) Since lens forms an upright and magnified image of an object, the lens used here is a convex lens....

(a) Name the kind of lens. (b) State the nature of the image whether real or virtual? Solution: (a) Since the lens forms an inverted image of an object, it is a convex lens. (b) The image...

Solution: Because the Sun is at infinity, the convex lens creates an image at the second focus point that is both real and greatly reduced in size. The sun's beams converge at the focus of the lens,...

Solution: The object is positioned between the focal point F1 and the convex lens, and its image is generated on the same enlarged side of the lens. As a result, this lens can be used as a...

Solution: The image of an object placed at the focal point of a concave lens is vertical and diminished.

Solution: The above figure shows the image formed is i) real, ii) enlarged and iii) inverted.

Solution: This ray diagram shows that a diverging lens cannot form a real image of an object placed anywhere on its principal axis.

Solution: Suppose that the object is placed beyond 2F1 and its diminished image is formed between F2 and 2F2. Such an image is real and inverted. Here the object is AB and the image is real and...

Solution: (a) Concave lens is a lens that forms the image between the object and itself. (b) This figure shows the ray diagram (c) The three characteristics of the image are (i) virtual, (ii) erect...

(a) Name the kind of lens. (b) Where is the object placed in relation to the lens? (c) Draw a ray diagram to show the formation of the image. (d) Name the device which uses this principle. Solution:...

Solution: (a) The object is placed at a distance equal to twice the focal length of the lens. (b) (c) The image formed is real and inverted

(a) Where is the object placed? (b) Draw a ray diagram to show the formation of an image. (c) State its three characteristics of the image. Solution: (a) Since the image of an object formed by a...

(i) Draw two rays to locate the position of the image. (ii) State the position of the image with reference to the lens. (iii) Describe the three characteristics of the image. (iv) Describe how the...

(a) Draw two rays from the object AB and complete the diagram to locate the position of the image CD. Also mark on the diagram the position of eye from where the image can be viewed.   (b) State...

(i)Name the lens LL’. (ii)What are the points O and O’ called? (iii)Complete the diagram to form the image of the object AB. (iv)State three characteristics of the image. Solution: (i) LL’...

(a) Name the lens LL’. (b) What are the points O and O’ called? (c) Complete the diagram to form the image of the object AB. (d) State the three characteristics of the image. ...

Solution: Real Image (i) The actual junction of the rays refracted by the lens produces a real image.(ii) On a screen, a real image can be obtained.(iii) In relation to the object, a real image is...

Solution: When a source of light is placed at the first focal point, which is the focal point to the left of the optical centre of the convex lens, the rays of light refracted through the convex...

Solution -

Solution: (i) A ray of light incident at the lens' optical centre O goes through the lens without deviating. (ii) After refraction, a ray of light incident parallel to the primary axis of the lens...

(a)An object is placed at a distance of more than 40 cm from a convex lens of focal length 20 cm. The image formed is real, inverted and……………. (b)An object is placed at a distance 2f from a convex...

Solution: According to the question, Object distance is u = – 8 cm Focal length is f = 10 cm We have to find out the image distance v The expression of the Len’s formula is 1 / f = 1 / v – 1 / u...

Solution: According to the question Object distance is u = -30 cm Focal length is f = -30 cm We have to find out the image distance v According to the Len’s formula, we have - 1 / f = 1 / v – 1 / u...

Solution: The concave lens' image is always imaginary, upright, and smaller than the object. As a result, the magnification is always positive and smaller than a factor of one. The relationship...

Solution: The image is real and inverted, as indicated by the negative magnification value. The magnification is larger than 1, implying that the image is magnified. As a result, the lens should be...

Solution: According to the question - The power of a lens is -2.0 D We know that the expression of Power of a lens is P = 1 / f Upon substituting values, we have => -2 = 1 / f Or, f = 1 / -2 = -0.5...

Solution: According to the question, Focal length is f = +25 cm = +0.25 m We know that the expression of power of a lens is - P = 1 / f Substituting value of focal length, we get => P = 1 / +0.25...

Solution: The expression for the power of lens is P = 1 / F According to the question, it is a concave lens so focal length will be negative. Upon substituting values, we get => P = 1 / - 0.2 m...

(a) The position of image, and (b) The focal length of the lens. Solution: According to the question, Distance of the object u is – 30 cm And the concave lens forms an erect image of 1/3...

(a) The position of image, and (b) The focal length of the lens Solution: (a) According to the question, the convex lens forms an inverted, real image of the same size as the object. So,...

(a) the position of the image, and (b) the magnification of the image Solution: According to the question, Object distance is u = – 20 cm Focal length is f = – 20 cm (concave lens) (a) Expression...

(i) the focal length of lens, (ii) the magnification of the image. Solution: According to the question, Object distance is u = – 45 cm Image distance is v = + 90 cm (a) Since the image is formed on...

(i) the focal length, (ii) the magnification, and (iii) the nature of image. Solution: According to the question, Object distance is u = – 15 cm Image distance is v = – 60 cm (i) We know that the...

Solution: According to the question, Image distance is v = 75 cm Object distance is u = -25 cm The expression for Lens formula is - 1 / v – 1 / u = 1 / f Upon substituting values, we get 1 / f = 1 /...

Solution: According to the question, Focal length of the given convex lens is f = + 0.12 m Magnification is m = -3 (real image) For a lens, expression of magnification is m = v / u upon substituting...

Solution: According to the question, Focal length is f = +25 cm Image is virtual and magnified, with magnification (m) = +2 For a lens, expression for the magnification is m = v / u Or, +2 = v / u...

(a) What is the nature of the image? (b) Find the focal length of the lens. Solution: According to the question, Object distance is u = -20 cm Image distance is v = – 10 cm (a) Because the image is...

(b) What is the focal length of the lens in part (a)? Solution: According to the question, Height of a candle is 3 cm Height of the image is 6 cm Image distance is 30 cm (a) Expression for the...

(a) 25 (b) 10 (c) 1 (d) 6 Solution: According to the question, Focal length of convex lens is f = +5 cm Expression for the magnifying power of the convex lens (simple microscope) is M = 1 + D / f...

(a) A real and diminished image (b) A real and magnified image (c) A virtual and magnified image (d) A virtual and diminished image Solution: The correct option is (c). Therefore, a magnifying glass...

(i) a distant object and (ii) a printed page? Solution: (i) If the inverted image of a distant object is seen through the lens, the lens is convex; if the upright image is visible, the lens is...

Solution: A vertical stand, a plane mirror, a lens, and a pin are required to determine the focal length using a plane mirror. Place the lens L on a plane mirror MM' horizontally. Arrange the pin P...

(i)Draw two rays from the point O of the object pin to show the formation of image I at O itself. (ii)What is the size of the image I? (iii)State two more characteristics of the image I. (iv)Name...

Solution: A convex lens' approximate focal length can be calculated using the premise that a beam of parallel rays incident from a distant object converges in the lens' focal plane. The 0 centimetre...

Solution: Some of the applications of a convex lens are (i) A telescope, camera, slide projector, or other device's objective lens is a convex lens that creates a real and inverted image of the...

Solution: Expression for the magnifying power of the microscope is as follows - M = 1 + D / f, where M denotes the magnifying power Where f denotes the lens' focal length and D denotes the shortest...

Solution: The item is positioned between the lens and the point of primary focus. The image is formed in the distance between the lens and the primary focus.

Solution: When the object is situated between focal length and the optical centre of a convex lens, the image of the object will form on the same side of the lens. In the diagram below, AB...

Solution: A convex lens with a short focal length is known as a magnifying glass. For practical use, it is fitted in a lens holder. It's used to see and read little characters and figures...

(a) Its first focus(b) Its optical entre(c) Its second focus(d)The centre of curvature of its second surface Solution: The ray of light after refraction passes through or appears to come from its...

(a) its optical centre(b) its first focus(c) its second focus (d) it’s centre of curvature of the first surface Solution: the correct option is (b). The incident ray passes through its first...

(a) convex of focal length 1.0 cm (b) convex of focal length 1.0 m (c) concave of focal length 1.0 cm (d) concave of focal length 1.0 m Solution: The correct option is (d). According to the...

(a) Decreases  (b) Increases  (c) Does not change  (d) First increases then decreases. Solution: The correct option is (b). On reducing the focal length of a lens, the power of the...

(a) 1 / u + 1 / v = 1 / f (b) 1 / u – 1 / v = 1 / f (c) 1 / v – 1 / u = 1 / f (d) f = (u + v) / uv Solution: The correct option is (c). Expression for the lens formula is - 1 / v – 1 / u = 1 / f...

(a) The lens is concave (b) The image is virtual (c) The image is magnified (d) The images are real and diminished formed by a convex Solution: The correct option is (d). The image is real and...

Solution: The power of a thick lens is more than a thin one because a thick lens has greater curvature or lesser focal length as compared to a thin lens.

Solution: We know that the expression for power of a lens is - P = 1/f, where f is the focal length of the lens Since the focal length of a convex lens is positive and it is negative for a concave...

Solution: The direction in which a light ray is deviated by the lens determines the power sign. It's possible that the power is positive or negative. The power of a lens is positive if it deviates a...

Solution: Power pf a lens is inversely proportional to its focal length in the following manner - P = 1/f where f is the focal length. So according to the expression, when the focal length is...

Solution: The ability of a lens to concentrate incoming rays is known as the power. Power of a lens is related to its focal length in the following manner - P = 1 / f Or, Power of lens (in D) = 1 /...

Solution: The ability of a lens to concentrate light rays falling on it is known as power of the lens. Expression for the Power of lens is as follows : P = 1 / f where f is the focal length. Dioptre...

Solution: (i) A positive magnification denotes that the image generated is virtual. Whereas a negative sign denotes that the image formed is genuine. (ii) A positive magnification denotes that the...

Solution: The ratio of the size of the image to the size of the object is known as the magnification. The expression for magnification is m = -v / u where v is the image distance and u is the object...

Solution: According to the Lens formula, we have - 1 / v – 1 / u = 1 / f where ‘u' is the object's distance from the lens' optical centre The image's distance from the optical center is v, and the...

In each case, state the kind of lens. Solution: (i) If a lens' focal length is positive, the lens is said to be convex.(ii) A lens is called a concave lens if its focal length is negative.

Solution: (a) The primary axis is the axis along which distances are measured. These distances are measured from the lens's optical centre.(b) Positive distances are those measured in the same...

(a) Virtual, inverted and diminished (b) Virtual, upright and diminished (c) Virtual, inverted and enlarged (d) Virtual, upright and enlarged Solution: Correct option is (b). A concave lens forms...

(a) Real and enlarged (b) Real and diminished (c) Virtual and enlarged (d) Virtual and diminished. Solution: The correct option is (c). The image is virtual and enlarged.

(a) 20 cm (b) 10 cm (c) 15 cm (d) 40 cm Solution: The correct option is (b) 10 cm, the focal length of a convex lens is 10 cm. Formula Used - 1/f = 1/v + 1/u According to the question, v = 20cm and...

(d)A ray of light incident at the optical centre of the lens, passes undeviated after refraction (e)A concave lens forms a magnified or diminished image depending on the distance of...

(a)A convex lens has a divergent action and a concave lens has a convergent action. (b)A concave lens, if kept at a proper distance from an object, can form its real image (c)A ray of light incident...

Solution: (a) (b)

(a)If half part of a convex lens is covered, the focal length __________ change, but the intensity of image _________. (b)A convex lens is placed in water. Its focal length will ________. (c)The...

Solution: (a) (b)

(i)Name the lens formed by the combination. (ii)What is the line XX’ called? (iii)Complete the ray diagram and show the path of the incident ray AB after passing through the lens. (iv)The final...

(i) convex lens, (ii) concave lens. Draw a diagram in each case to show the refraction of light through the lens. Solution: (i) A convex lens refracts an oblique parallel beam. (ii) A concave lens...

(i) a lens has both its focal lengths equal. (ii) a ray passes undeviated through the lens. Solution: (i) If the focal lengths on both sides of the lens are equal, medium is the...

Solution: The focal plane of a lens is a plane parallel to the primary axis that passes across the focus.

Solution: The focal length of a lens is the distance of focus point from the optical centre of the lens.

Solution: This point F appears to come from a second focus

Solution: This point F is known as the second focus.

(a) Draw a ray diagram to show the incident ray and its corresponding emergent ray. (b) The incident ray when produced meets the principal axis at a point F. Name the point F. Solution: (a) (b) When...

Solution:

Solution:

Solution: From either side, a light ray can travel through a lens. As a result, a lens has two primary foci.The first focal point of a concave lens is a point F1 on the principal axis of the lens...

Solution: From either side, a light ray can travel through a lens. As a result, a lens has two primary foci, one on each side.The first focal point of a convex lens is a point F1 on the primary axis...

Solution: The lens is called equiconvex or equi-concave if the radius of curvature of both surfaces are equal.

(a) What special name is given to this point on the principal axis? (b) Draw a labelled diagram to support the answer in part(a). Solution: (a) Optical centre is the name of this point on the...

Solution: The optical centre is a position on the lens's primary axis where a ray of light travelling through it emerges parallel to its incidence direction. In the diagram below, the letter O...

Solution: A line connecting the centres of curvature of the two surfaces of the lens is known as the principal axis.

Solution: When a parallel light beam strikes a convex lens, the prism in the upper portion bends the incident ray downwards. Similarly, the prism in the bottom section bends it upwards. The ray...

Solution: The concave lens, which has two glass prisms and one glass slab, is shown above. One glass prism is located above the glass slab, while the other is located below it.

Solution: The convex lens, which has two glass prisms and one glass slab, is depicted in the diagram above. One glass prism is positioned above the glass slab, while the other is positioned below...

Solution: Concave lens are known to show the divergent action on a light beam. The diagram below shows this behaviour of concave lens -

(i) an equiconcave lens or an equiconvex lens? (ii) a concavo-convex lens or a convexo-concave lens? Solution: (i) A converging equiconvex lens(ii) A converging lens is a...

(a) appearance, and (b) action on the incident light.  Solution: Convex Lens : (i) It is thick in the center and thin on the outside.(ii) The incident rays are converged towards the primary...

Solution: There are two kinds of lenses. (i) Convex or converging lens and (ii) Concave or diverging lens Basic diagrams of these lenses are -

Solution: ‘A transparent refracting medium confined by either two spherical surfaces or one spherical surface and another surface plane' is known as a Lens.