Explanation
It is given that,
R1=−3cm
Focal length of concave lens f1=−32
R2=2cm
Focal length of convex lens f2=1cm
Combined focal length is
1f=1f1+1f2
1f=−23+1
1f=13........(1)
The relation between redfractive index and focal length is
1f=(μ−1)[1R1−1R2]
13=(μ−1)[1−3−12]
13=(μ−1)[−56]
μ=35
Total deviation for yellow ray produced
δy=δcy−δfy+δcy
=2δcy−δfy
=2(μcy−1)A−(μ−cy−1)A1
Angular dispersion,
δv−δr=[(μvc−1)A−(μvf−1)A1+(μvc−1)A]
−[(μrc−1)A−(μrf−1)A1+(μr−1)A]
2(μvc−1)A−(μvf−1)A1
For net angular dispersion to be zero,
δr−δr=0
2(μvc−1)A=(μvf−1)A1
If μ, represents the refractive index when a light ray goes from medium i to j, then 2μ1 x 3μ2 x 4μ3 is equal to
A light ray is incident normally on the surface AB of a prism of refracting angle 60∘. 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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