Explanation
An observer looks at a distant treeof height $$10$$ m with atelescope of magnifying power of $$20$$. To theobserver the tree appears :
.
The condition for minimum deviation in a prism is given as
1. $$i = e$$ where $$i$$ is the angle of incidence and $$e$$ is the angle of emergence.
2. $${r_1} = {r_2}$$ where $${r_1}$$ and $${r_2}$$ are the angle of refraction of the two refracting surfaces of the prism.
Hence the first refracted ray inside the prism should be parallel to the base of the prism.
In the figure we can see that the above condition is satisfied by the prisms $$A$$ and $$B$$.
Hence the correct answer is option (C).
A $$10 \mathrm{mm}$$ long awl pin is placed vertically in front of a concave mirror. A $$5 \mathrm{mm}$$ long image of the awl pin is formed at $$30 \mathrm{cm}$$ in front of the mirror. The focal length of this mirror is:
Given: object size $$\mathrm{h}=10\mathrm{mm},$$ image size $$\mathrm{h}^{\prime}=5 \mathrm{mm},$$ image distance$$\mathrm{v}=-30 \mathrm{cm},$$
To find object distance=? $$\mathrm{f}=?$$
from the magnification formula $$=\dfrac{h^{\prime}}{h}=\dfrac{-v}{u}$$
or $$\dfrac{5m m}{10 m m}=\dfrac{-30 c m}{u}$$ or $$\dfrac{-30 c m}{u}=\dfrac{1}{2}$$
or $$u=-60$$
Now, it can be calculated as follows:
$$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$$
or $$\dfrac{1}{-30\mathrm{cm}}+\dfrac{1}{-60 \mathrm{cm}}=\dfrac{1}{f}$$ or $$\dfrac{-2-1}{60}=\dfrac{1}{f}$$
$$\dfrac{1}{f}=-\dfrac{1}{20\mathrm{cm}}$$ or $$f=-20 \mathrm{cm}$$ (the negative sign confirms that the calculated focal length is of a concavemirror)
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