✓ Board-verified
Science · 2026 · 4 marks
CBSE 2026 · Region 1 · Set 1 · Q38
Read the following passage and answer the questions given: Lenses can form different types of images depending upon their focal length and position…
Read the following passage and answer the questions given : Lenses can form different types of images depending upon their focal length and position of object. A convex lens can create real, inverted or virtual, erect images, while a concave lens forms only virtual and diminished images. The focal length determines the power of lens. Convex lenses have positive focal length while concave lenses have negative focal length by convention. When lenses are placed together, their combined power is determined by the sum of their individual powers. Ray diagrams help to visualize how light converges or diverges through lens to form an image.
(a)
A convex lens of focal length $\displaystyle 20$ cm is used to form an image. If an object is placed at $\displaystyle 40$ cm from the lens, what will be the position and nature of image ?
(b)
Illustrate the formation of image with the help of ray diagram, when the object is placed between the optical centre and principal focus of concave lens.
(c)
(i)
A lens combination consists of a convex lens of focal length $\displaystyle 30$ cm and a concave lens of focal length $\displaystyle 15$ cm placed together. Find the equivalent focal length and power of this lens combination.
OR
(ii)
Two lenses are placed in contact. One is a concave lens with focal length $\displaystyle 2$ m and the other is a convex lens with focal length $\displaystyle 1.5$ m . What type of lens will the combination behave as (convex or concave) ? Give reason.
Marking-scheme solution
(a)
Position - Image will form at $\displaystyle 40$ cm / 2F /C
Nature - Real and inverted
Alternate answer
\[\mathrm{f}=+20 \mathrm{~cm}, \mathrm{u}=-40 \mathrm{~cm}
\]
Using Lens Formula,
\[\begin{array}{l}
\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\
\frac{1}{v}-\frac{1}{-40}=\frac{1}{20} \\
\frac{1}{v}=\frac{1}{20}+\frac{1}{-40} \\
v=+40 \mathrm{~cm}
\end{array}
\]
Position - $\displaystyle 40$ cm on the other side of the lens
Nature - Real and inverted
(b)
(c)
\[\begin{array}{l}
\mathrm{f}_{1}=30 \mathrm{~cm}=0.3 \mathrm{~m}, \quad \mathrm{f}_{2}=-15 \mathrm{~cm}=-0.15 \mathrm{~m} \\
\mathrm{P}=\frac{1}{\mathrm{f}} \\
\mathrm{P}_{1}=\frac{+1}{0.3} \mathrm{D} ; \mathrm{P}_{2}=\frac{-1}{0.15} \mathrm{D}
\end{array}
\]
Equivalent power, $\displaystyle \mathrm{P}=\mathrm{P}_{1}+\mathrm{P}_{2}$
\[\mathrm{P}=-3.33 \mathrm{D}
\]
Equivalent focal length, $\displaystyle \mathrm{f}=\frac{1}{\mathrm{P}}$
\[\mathrm{f}=\frac{-1}{3.33}=-0.3 \mathrm{~m}=-30 \mathrm{~cm}
\]
Combination Lens will behave like convex lens
$\displaystyle \mathrm{f}_{1}=-2 \mathrm{~m}, \mathrm{f}_{2}=1.5 \mathrm{~m}$
\[\begin{array}{l}
\mathrm{P}=\frac{1}{\mathrm{f}} \\
\mathrm{P}_{1}=\frac{-1}{2} \mathrm{D}, \mathrm{P}_{2}=\frac{+1}{1.5} \mathrm{D} \\
\mathrm{P}=\mathrm{P}_{1}+\mathrm{P}_{2} \\
\mathrm{P}=\frac{1}{6} \\
\mathrm{f}=+6 \mathrm{~m}
\end{array}
\]
The focal length of combination is positive.
Alternate answer
Combination Lens will behave like convex lens.
Convex lens - Less ' f ', More ' $\displaystyle \mathrm{P}_{1}$ '
Concave lens - More 'f', Less ' $\displaystyle \mathrm{P}_{2}$ '
Combined Power $\displaystyle =\mathrm{P}_{1}+\mathrm{P}_{2}$, which will be positive.
All Light - Reflection and Refraction questions2026 question paperNCERT solutions: Light - Reflection and RefractionPractice Light - Reflection and Refraction →
More from Light - Reflection and Refraction
- With the help of above given ray diagram, calculate the angle of refraction. Given that the refractive index of Medium 2 with respect to Medium 1 is…2026
- A student wants to obtain an erect image of an object using a concave mirror of 10 cm focal length. What will be the distance of the object from…2023
- 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…2023
- If the absolute refractive indices of two media X and Y are 6/5 and 4/3 respectively, then the refractive index of Y with respect to X will be:2025
- A student wants to project the image of a candle flame on a screen 60 cm in front of a mirror by keeping the candle at a distance of 15 cm from its…2023
- An object is placed in front of a concave mirror of focal length 12 cm. If distance of the object from the pole of the mirror is 8 cm, then use…2024
- (i) "In refraction of light through a rectangular glass slab, the emergent ray is always parallel to the direction of the incident ray." Why? Explain…2025
- The above ray diagram is to show image formation by a reflecting telescope. Reflecting telescope revolutionized our way of looking into the sky.…2026
CBSE Class 10 Science past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.