Board question: The power of a lens is +4D. Find the focal length of this lens. An object is placed at a distance of 50 cm from the optical centre of this lens. State the nature and magnification of the image formed by the lens and also draw a ray diagram to justify your answer.
Paper: CBSE 2023 Main 31/4/1 Q31 (3 marks).
What the question asks: find the focal length from the power, then find where the image forms, say what kind of image it is and how big, and draw the ray diagram.
Key words: plus power means a convex (converging) lens. Distances are measured from the optical centre (the middle of the lens) along the principal axis (the line through the middle of the lens).
Step 1, focal length: P = 1/f (spoken “1 by f”), with f in metres, because 1 dioptre is the power of a lens whose focal length is 1 metre. So f = 1/(+4) m = +0.25 m = +25 cm (plus: a convex lens).
Step 2, signs: distances measured against the direction of the light are negative. The object is on the side the light comes from, so u = −50 cm.
Step 3, lens formula: 1/v − 1/u = 1/f, where v is the image distance. So 1/v = 1/f + 1/u = 1/25 + 1/(−50) = 1/50, and v = +50 cm.
Step 4, nature and size: v is positive, so the rays really meet on the other side of the lens: the image is real and inverted. Magnification m = v/u = 50/(−50) = −1: the minus sign means inverted, and 1 means the same size.
Step 5, ray diagram: 50 cm is twice 25 cm, so the object is at 2F. A ray parallel to the principal axis bends through the focus; a ray through the optical centre goes straight on. They meet at 2F on the other side. Draw arrows on the rays to show their direction.
Answer, marking scheme (CBSE 2023 Main 31/4/1 Q31): f (m) = 1/P (½); f = 0.25 m = 25 cm (½); real and inverted (½); magnification = −1 (½); ray diagram (1; ½ is deducted if the ray directions are not marked).
The trap: taking u as +50 cm. That gives the wrong v. The object is against the light, so u = −50 cm.
Quick recall (Light): Mirror: radius R and focal length f? → R = 2f
Comment challenge (CBSE 2023 Main 31/4/2 Q30(A)): An object is kept at a distance of 1m from a lens of power +2D : (i) Identify the type of lens. (ii) Calculate its focal length and distance of the image formed. Answer in the pinned comment.
References: NCERT Class 10 Science, Ch 9 Light – Reflection and Refraction, §9.3 Refraction by spherical lenses (image formation by a convex lens: object at 2F1 gives a real, inverted, same-size image at 2F2; New Cartesian sign convention; lens formula 1/v − 1/u = 1/f; magnification m = v/u; power P = 1/f with f in metres, unit dioptre); CBSE Class 10 board questions and marking schemes.
Narration uses a licensed AI voice (Azure AI Speech, Indian English). Script and physics research by ScoreSpark.
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Board question (CBSE 2023 Main 31/4/1 Q31, 3 marks): The power of a lens is +4D. Find the focal length of this lens. An object is placed at a distance of 50 cm from the optical centre of this lens. State the nature and magnification of the image formed by the lens and also draw a ray diagram to justify your answer.
Board answer (CBSE marking scheme 2023 Main 31/4/1 Q31): f (m) = 1/P (½); f = 1/(+4 D) = 0.25 m = 25 cm (½); real and inverted (½); magnification = −1 (½); ray diagram (1; ½ is deducted if the ray directions are not marked).
Comment challenge (CBSE 2023 Main 31/4/2 Q30(A)): An object is kept at a distance of 1m from a lens of power +2D : (i) Identify the type of lens. (ii) Calculate its focal length and distance of the image formed.
Answer: (i) Power is positive, so it is a convex (converging) lens. (ii) f = 1/P = 1/(+2 D) = 0.5 m = 50 cm. With u = −100 cm: 1/v − 1/u = 1/f gives 1/v = 1/50 − 1/100 = 1/100, so v = +100 cm = 1 m (a real image 1 m behind the lens; the object is at 2F, so the image is at 2F on the other side). CBSE 2023 marking scheme (31/4/2 Q30(A)): convex (1); f = 1/P (½); f = 0.5 m = 50 cm (½); 1/f = 1/v − 1/u (½); v = 100 cm = 1 m (½).
Quick recall (Light): Mirror: radius R and focal length f? → R = 2f