- Draw a ray diagram to show image formation when the concave mirror produces a real, inverted and magnified image of the object.
- Obtain the mirror formula and write the expression for the linear magnification.
- Explain two advantages of a reflecting telescope over a refracting telescope.
CBSE DELHI - OUTSIDE DELHI - FOREIGN SET 2 2018
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- Object between F and C. The image is Beyond C.
- No chromatic aberration.
- Spherical aberration reduces significantly.
- Low set up cost, lighter, forms brighter image.
From the concavemirror.
Using cartesian sign convention, we find
Now, $\triangle\text{A}'\text{B}'\text{C}\sim\triangle\text{ABC}$Object distance,BP = -uImage distance,B'P = -vFocal length,FP = -fRadius of curvature,CP = -R = -2f
$\therefore\ \frac{\text{A}'\text{B}'}{\text{AB}}=\frac{\text{CB}'}{\text{BC}}=\frac{\text{CP}-\text{B}'\text{P}}{\text{BP}-\text{CP}}=\frac{-\text{R}+\text{v}}{-\text{u}+\text{R}}\ ....(1)$
As $\angle\text{A}'\text{PB}'=\angle\text{APB},$ therefore,
$\triangle\text{A}'\text{B}'\text{P}\sim\triangle\text{ABP}.$
Consequently,
$\frac{\text{A}'\text{B}'}{\text{AB}}=\frac{\text{B}'\text{P}}{\text{BP}}=\frac{-\text{v}}{-\text{u}}=\frac{\text{v}}{\text{u}}\ ....(2)$
From equations (1) and (2), we get
$\frac{-\text{R}+\text{v}}{-\text{u}+\text{R}}=\frac{\text{v}}{\text{u}}$
or, $-\text{uR}+\text{uv}=\text{-uv}+\text{vR}$
or, $\text{vR}+\text{uR}=2\text{uv}$
Dividing both sides by uvR, we get
$\frac{1}{\text{u}}+\frac{1}{\text{v}}=\frac{2}{\text{R}}$
But, R = 2f
$\frac{1}{\text{u}}+\frac{1}{\text{v}}=\frac{1}{\text{f}}$
This proves themirror formula for a concavemirror, when it forms a real image.
Linear magnification is given by:
$\text{m}=\frac{\text{h}_2}{\text{h}_1}=\frac{-\text{v}}{\text{u}}$
here $h_2, h_1$ are heights of image and object respectively v, u are distances of image and object from the pole of mirror respectively.- Nature Real, Inverted and Larger than object.
- Consider an object AB placed on the principal axis beyond the centre of curvature C of a concave mirror of small aperture, as shown in Fig. Aray AM from the object travels parallel to the principal axis and after reflection from the mirror it passes through focus F. Another ray AP is incident on the pole P of the mirror and is reflected along PA' in accordance with the laws of reflection so that $\angle\text{APB}=\angle\text{BPA}'.$ The two reflected rays meet at point A'. Thus A' is the real image of A. The image of any point on AB will lie on a corresponding point of A'B. Hence A'B is the real image of AB formed by reflection.
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