✓ Board-verified
Science · 2023 · 3 marks
CBSE 2023 · Region 5 · Set 3 · Q29
A student has focussed the image of an object of height $\displaystyle 3$ cm on a white screen using a concave mirror of focal length $\displaystyle 12$ cm. If the distance of the object from the mirror is $\displaystyle 18$ cm, find the values of the following :(i)Distance of the image from the mirror(ii)Height of the imageDefine power of a lens. The focal length of a lens is -$\displaystyle 10$ cm. Write the nature of the lens and find its power. If an object is placed at a distance of $\displaystyle 20$ cm from the optical centre of this lens, according to the New Cartesian Sign Convention, what will be the sign of magnification in this case ?
A student has focussed the image of an object of height $\displaystyle 3$ cm on a white screen using a concave mirror of focal length $\displaystyle 12$ cm. If the distance of the object from the mirror is $\displaystyle 18$ cm, find the values of the following :
(i)
Distance of the image from the mirror
(ii)
Height of the image
Define power of a lens. The focal length of a lens is -$\displaystyle 10$ cm. Write the nature of the lens and find its power. If an object is placed at a distance of $\displaystyle 20$ cm from the optical centre of this lens, according to the New Cartesian Sign Convention, what will be the sign of magnification in this case ?
Marking-scheme solution
(a) \(\displaystyle \mathrm{h}_{1}=3 \mathrm{~cm}\)
\[\begin{aligned}
& f=-12 \mathrm{~cm} \\
& u=-18 \mathrm{~cm}
\end{aligned}
\]
(i) Image Distance
\[\begin{aligned}
& \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\
& \frac{1}{-12}=\frac{1}{v}+\frac{1}{-18} \\
& \frac{1}{-12}+\frac{1}{18}=\frac{1}{v} \\
& \frac{1}{v}=\frac{-3+2}{36}=\frac{-1}{36} \\
& v=-36 \mathrm{~cm}
\end{aligned}
\]
(ii)
Height of image
\[\begin{aligned}
& \mathrm{m}=\frac{\mathrm{h}_{2}}{\mathrm{~h}_{1}}=\frac{-\mathrm{v}}{\mathrm{u}} \\
& \frac{\mathrm{~h}_{2}}{3}=\frac{-[-36]}{-18} \\
& \mathrm{~h}_{2}=\frac{-36}{18} \times 3 \\
& \mathrm{~h}_{2}=-6 \mathrm{~cm}
\end{aligned}
\]
Degree of convergence or divergence of light/ Reciprocal of focal length of lens in metre.
It is diverging/concave lens
\(\displaystyle \mathrm{P}=\frac{1}{f(m)}=\frac{100}{f(c m)}\)
\[P=\frac{100}{-10 \mathrm{~cm}}=-10 \mathrm{D}
\]
Sign of magnification = + or positive
More from Light - Reflection and Refraction
- If the absolute refractive indices of two media X and Y are 6/5 and 4/3 respectively, then the refractive…2025 · asked 3×
- (i) "In refraction of light through a rectangular glass slab, the emergent ray is always parallel to the…2025 · asked 3×
- The above ray diagram is to show image formation by a reflecting telescope. Reflecting telescope…2026 · asked 3×
- Draw ray diagrams to show the nature, position and relative size of the image formed by a convex mirror when…2025 · asked 3×
- A highly polished surface such as a mirror reflects most of the light falling on it. In our daily life we use…2024 · asked 3×
- Out of the two lenses, one concave and the other convex, state which one will diverge a parallel beam of…2025 · asked 3×
- The power of a lens is -0.25 D. Based on this information, find out The type of lens and its focal length.…2025 · asked 3×
- Study the data given below showing the focal length of three concave mirrors A, B and C and the respective…2024 · asked 3×
CBSE Class 10 Science past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.