Explanation
It is given that,
$${{R}_{1}}=-3\,cm$$
Focal length of concave lens $${{f}_{1}}=\dfrac{-3}{2}$$
$${{R}_{2}}=2\,cm$$
Focal length of convex lens $${{f}_{2}}=1\,cm$$
Combined focal length is
$$ \dfrac{1}{f}=\dfrac{1}{{{f}_{1}}}+\dfrac{1}{{{f}_{2}}} $$
$$ \dfrac{1}{f}=\dfrac{-2}{3}+1 $$
$$ \dfrac{1}{f}=\dfrac{1}{3}........(1) $$
The relation between redfractive index and focal length is
$$ \dfrac{1}{f}=\left( \mu -1 \right)\left[ \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right] $$
$$ \dfrac{1}{3}=\left( \mu -1 \right)\left[ \dfrac{1}{-3}-\dfrac{1}{2} \right] $$
$$ \dfrac{1}{3}=\left( \mu -1 \right)\left[ \dfrac{-5}{6} \right] $$
$$ \mu =\dfrac{3}{5} $$
Total deviation for yellow ray produced
$$\delta y=\delta_{cy}-\delta_{fy}+\delta _{cy}$$
$$=2\delta_{cy}-\delta_{fy}$$
$$=2(\mu_{cy-1})A-(\mu_{-cy}-1)A^1$$
Angular dispersion,
$$\delta_v-\delta_r=[(\mu_{vc}-1)A-(\mu_{vf}-1)A^1+(\mu_{vc}-1)A]$$
$$-[(\mu_{rc}-1)A-(\mu_{rf}-1)A^1+(\mu_{r}-1)A]$$
$$2(\mu_{vc}-1)A-(\mu_{vf}-1)A^1$$
For net angular dispersion to be zero,
$$\delta_r-\delta_r=0$$
$$2(\mu_{vc}-1)A=(\mu_{vf}-1)A^1$$
If $$\mu ,$$ represents the refractive index when a light ray goes from medium i to j, then $$_2{\mu _1}{\text{ x}}{{\text{ }}_3}{\mu _2}{\text{ x}}{{\text{ }}_4}{\mu _3}$$ is equal to
A light ray is incident normally on the surface $$AB$$ of a prism of refracting angle $$60^\circ$$. If the light ray does not emerge from $$AC$$, then find the refractive index of the prism.
Given that,
Diameter $$d=2\,m$$
Wave length $$\lambda =5000\,\overset{\circ }{\mathop{A}}\,$$
Now, minimum angular separation is
$$ \Delta \theta =\dfrac{1.22\lambda }{d} $$
$$ \Delta \theta =\dfrac{1.22\times 5000\times {{10}^{-10}}}{2} $$
$$ \Delta \theta =0.3\times {{10}^{-6}}\,rad $$
Hence, the resolving power is $$0.3\times {{10}^{-6}}\,rad$$
Given,
Radius of curvature, $$R$$
Refractive index = $$ 1.5 $$
Focal length of lens, $$f=\dfrac{R}{(\mu-1)}=\dfrac{R}{(1.5-1)}=2R$$
Magnification from mirror
$$m = \dfrac { h _ { i } } { h _ { o } } = \dfrac { - v } { u }$$
Whereas
$$h _ { i } =$$ Height of image from principal axis
$$h _ { o } =$$ Height of object from principal axis
$$u =$$ Object distance
$$v =$$ Image distance
Angular magnification $$m = \dfrac { f _ { o } } { f _ { e } } = \dfrac { 150 } { 5 } = 30$$
$$\therefore \quad \dfrac { \tan \beta } { \tan \alpha } = 30 \Rightarrow \tan \beta = \tan \alpha \times 30$$
$$\tan \beta = \left( \dfrac { 50 } { 1000 } \times 30 \right) = \dfrac { 15 } { 10 } = \dfrac { 3 } { 2 }$$
$$\tan \beta = \dfrac { 3 } { 2 } \Rightarrow \tan \beta = \tan ^ { - 1 } \left( \dfrac { 3 } { 2 } \right)$$
$$\theta = \beta \simeq 60 ^ { \circ }$$
At what angle will a ray of light be inclined on one face of an equilateral prism, so that the emergen ray may graze the second surface of the prism $$\left( {\mu = 2} \right)$$
Refractive indexes for, glass $${{\mu }_{g}}=1.5$$, water $${{\mu }_{w}}=\dfrac{4}{3}$$ and $${{\mu }_{a}}=1$$ .
If f is the focal length of the lens in air then
$$\dfrac{1}{{{f}_{a}}}=\left( \dfrac{{{\mu }_{g}}}{{{\mu }_{a}}}-1 \right)\times \left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)\,\,.........\,\,(1)$$
If f is the focal length of the lens in water then $$\dfrac{1}{{{f}_{w}}}=\left( \dfrac{{{\mu }_{g}}}{{{\mu }_{w}}}-1 \right)\times \left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)\,\,.........\,\,(2)$$
Divide equation (1) by (2)
$$\dfrac{\dfrac{1}{{{f}_{a}}}}{\dfrac{1}{{{f}_{w}}}}=\dfrac{\left( \dfrac{{{\mu }_{g}}}{{{\mu }_{a}}}-1 \right)}{\left( \dfrac{{{\mu }_{g}}}{{{\mu }_{w}}}-1 \right)}$$
$$ {{f}_{w}}=\left( \dfrac{{{\mu }_{g}}-{{\mu }_{a}}}{{{\mu }_{a}}}\times \dfrac{{{\mu }_{w}}}{{{\mu }_{g}}-{{\mu }_{w}}} \right){{f}_{a}} $$
$$ {{f}_{w}}=\left( \dfrac{1.5-1}{1}\times \dfrac{4/3}{1.5-4/3} \right)\times 20=80\,cm $$
Hence, focal length in water, $${{f}_{w}}=80\,cm$$
Focal length $$f=10\,cm$$
Refractive index $${{\mu }_{1}}=1.5$$
Refractive index $${{\mu }_{2}}=1.25$$
We know that,
$$ \dfrac{1}{f}=\left( n-1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right) $$
$$ \dfrac{1}{10}=\left( 1.5-1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)....(I) $$
$$ \dfrac{1}{f}=\left( 1.25-1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)....(II) $$
Now, divided equation (I) by equation (II)
$$ \dfrac{f}{10}=\dfrac{0.5}{0.25} $$
$$ f=20\,cm $$
Hence, the focal length is $$20\ cm$$
A convex lens produces an image of an object on a screen with a magnification of $$\frac{1}{2}$$ .When the lens is moved 30 cm away from the object, the magnification of the image is 2.The focal length of the lens is
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