Explanation:

Rays starting from \(O\) will suffer single refraction from spherical surface \(A P B\). Therefore, applying

\(\begin{aligned}& \dfrac{\mu_{2}}{v}-\dfrac{\mu_{1}}{u}=\dfrac{\mu_{2}-\mu_{1}}{R} ; \dfrac{1.0}{v}-\dfrac{2.0}{-15}=\dfrac{1.0-2.0}{-10} \\& =\dfrac{1}{v}=\dfrac{1}{10}-\dfrac{1}{7.5} \text { or } v=-30 {~cm}\end{aligned}\)
Therefore, image of \(O\) will be formed at 30 \(cm\) to the right of \(P\). Note that image will be virtual. There will be no effect of the surface \(C E D\).

Explanation:

Lenses have varying refractive indices for different wavelengths of light, with the refractive index decreasing as the wavelength increases. So, red light undergoes the least bending, while violet light undergoes the most.
For thickness of lews, there will be additional term added,

\(\frac{1}{f} = (\mu - 1)\left[ {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}} + \frac{{(\mu - 1)t}}{{\mu {R_1}{R_2}}}} \right]\)
\({f_{thick{\rm{ }}}} < {f_{thin{\rm{ }}}}\)
'Chromatic' means, we shall take 'colors' into account.
We know \(\lambda_{\text {red }}>\lambda_{\text {violet }}\)
\(\mu=\dfrac{c}{v}=\dfrac{c}{f \lambda} \Rightarrow \mu_{\text {red }} < \mu_{\text {violet }}\)
\(\Rightarrow f_{\text {red }}>f_{\text {violet }}\)
As they converge at different points, it is called 'aberration'
\(\Rightarrow\) Thick lens shows more aberration
\(\Rightarrow\) Assertion is true and reason is false.
So, correct option is (3).

Explanation:

Explanation:

Denote \(v_{O L}=\) Speed of object.
\(v_{I L}=\) Speed of image.
\(\Rightarrow v_{i L}=\left(\dfrac{v^{2}}{u^{2}}\right) v_{O L}\)
\(\Rightarrow 1=\dfrac{v^{2}}{u^{2}} \times 1\)
\(\Rightarrow v^{2}=u^{2}\)
We know, for lens:
\(\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\)
\(\Rightarrow v=\dfrac{u f}{u-f}\)
\(\Rightarrow\left(\dfrac{u f}{u-f}\right)^{2}=u^{2}\) [using equation (1)]
\( \Rightarrow u = 2f = 20\;cm\).
Correct option is (1).

Explanation:

\(\left( {A - R} \right)F = \frac{1}{P} = \frac{1}{1} = + 100\,cm\)
\(\left( {B - P} \right)\frac{1}{F} = \frac{1}{{0.5}} + \frac{1}{{0.5}} = + 400\,cm\)
\({\left( {C - S} \right)F = - \frac{1}{{2.5}} = - 0.4 = - 40\;cm}\)
\(\left( {D - Q} \right)\frac{1}{F} = \frac{1}{2} - \frac{1}{4} = 0.25 = 25\,cm\)
Option (2) is correct

Explanation:

Rays starting from \(O\) will suffer single refraction from spherical surface \(A P B\). Therefore, applying

\(\begin{aligned}& \dfrac{\mu_{2}}{v}-\dfrac{\mu_{1}}{u}=\dfrac{\mu_{2}-\mu_{1}}{R} ; \dfrac{1.0}{v}-\dfrac{2.0}{-15}=\dfrac{1.0-2.0}{-10} \\& =\dfrac{1}{v}=\dfrac{1}{10}-\dfrac{1}{7.5} \text { or } v=-30 {~cm}\end{aligned}\)
Therefore, image of \(O\) will be formed at 30 \(cm\) to the right of \(P\). Note that image will be virtual. There will be no effect of the surface \(C E D\).

Explanation:

Lenses have varying refractive indices for different wavelengths of light, with the refractive index decreasing as the wavelength increases. So, red light undergoes the least bending, while violet light undergoes the most.
For thickness of lews, there will be additional term added,

\(\frac{1}{f} = (\mu - 1)\left[ {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}} + \frac{{(\mu - 1)t}}{{\mu {R_1}{R_2}}}} \right]\)
\({f_{thick{\rm{ }}}} < {f_{thin{\rm{ }}}}\)
'Chromatic' means, we shall take 'colors' into account.
We know \(\lambda_{\text {red }}>\lambda_{\text {violet }}\)
\(\mu=\dfrac{c}{v}=\dfrac{c}{f \lambda} \Rightarrow \mu_{\text {red }} < \mu_{\text {violet }}\)
\(\Rightarrow f_{\text {red }}>f_{\text {violet }}\)
As they converge at different points, it is called 'aberration'
\(\Rightarrow\) Thick lens shows more aberration
\(\Rightarrow\) Assertion is true and reason is false.
So, correct option is (3).

Explanation:

Explanation:

Denote \(v_{O L}=\) Speed of object.
\(v_{I L}=\) Speed of image.
\(\Rightarrow v_{i L}=\left(\dfrac{v^{2}}{u^{2}}\right) v_{O L}\)
\(\Rightarrow 1=\dfrac{v^{2}}{u^{2}} \times 1\)
\(\Rightarrow v^{2}=u^{2}\)
We know, for lens:
\(\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\)
\(\Rightarrow v=\dfrac{u f}{u-f}\)
\(\Rightarrow\left(\dfrac{u f}{u-f}\right)^{2}=u^{2}\) [using equation (1)]
\( \Rightarrow u = 2f = 20\;cm\).
Correct option is (1).

Explanation:

\(\left( {A - R} \right)F = \frac{1}{P} = \frac{1}{1} = + 100\,cm\)
\(\left( {B - P} \right)\frac{1}{F} = \frac{1}{{0.5}} + \frac{1}{{0.5}} = + 400\,cm\)
\({\left( {C - S} \right)F = - \frac{1}{{2.5}} = - 0.4 = - 40\;cm}\)
\(\left( {D - Q} \right)\frac{1}{F} = \frac{1}{2} - \frac{1}{4} = 0.25 = 25\,cm\)
Option (2) is correct

Explanation:

Rays starting from \(O\) will suffer single refraction from spherical surface \(A P B\). Therefore, applying

\(\begin{aligned}& \dfrac{\mu_{2}}{v}-\dfrac{\mu_{1}}{u}=\dfrac{\mu_{2}-\mu_{1}}{R} ; \dfrac{1.0}{v}-\dfrac{2.0}{-15}=\dfrac{1.0-2.0}{-10} \\& =\dfrac{1}{v}=\dfrac{1}{10}-\dfrac{1}{7.5} \text { or } v=-30 {~cm}\end{aligned}\)
Therefore, image of \(O\) will be formed at 30 \(cm\) to the right of \(P\). Note that image will be virtual. There will be no effect of the surface \(C E D\).

Explanation:

Lenses have varying refractive indices for different wavelengths of light, with the refractive index decreasing as the wavelength increases. So, red light undergoes the least bending, while violet light undergoes the most.
For thickness of lews, there will be additional term added,

\(\frac{1}{f} = (\mu - 1)\left[ {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}} + \frac{{(\mu - 1)t}}{{\mu {R_1}{R_2}}}} \right]\)
\({f_{thick{\rm{ }}}} < {f_{thin{\rm{ }}}}\)
'Chromatic' means, we shall take 'colors' into account.
We know \(\lambda_{\text {red }}>\lambda_{\text {violet }}\)
\(\mu=\dfrac{c}{v}=\dfrac{c}{f \lambda} \Rightarrow \mu_{\text {red }} < \mu_{\text {violet }}\)
\(\Rightarrow f_{\text {red }}>f_{\text {violet }}\)
As they converge at different points, it is called 'aberration'
\(\Rightarrow\) Thick lens shows more aberration
\(\Rightarrow\) Assertion is true and reason is false.
So, correct option is (3).

Explanation:

Explanation:

Denote \(v_{O L}=\) Speed of object.
\(v_{I L}=\) Speed of image.
\(\Rightarrow v_{i L}=\left(\dfrac{v^{2}}{u^{2}}\right) v_{O L}\)
\(\Rightarrow 1=\dfrac{v^{2}}{u^{2}} \times 1\)
\(\Rightarrow v^{2}=u^{2}\)
We know, for lens:
\(\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\)
\(\Rightarrow v=\dfrac{u f}{u-f}\)
\(\Rightarrow\left(\dfrac{u f}{u-f}\right)^{2}=u^{2}\) [using equation (1)]
\( \Rightarrow u = 2f = 20\;cm\).
Correct option is (1).

Explanation:

\(\left( {A - R} \right)F = \frac{1}{P} = \frac{1}{1} = + 100\,cm\)
\(\left( {B - P} \right)\frac{1}{F} = \frac{1}{{0.5}} + \frac{1}{{0.5}} = + 400\,cm\)
\({\left( {C - S} \right)F = - \frac{1}{{2.5}} = - 0.4 = - 40\;cm}\)
\(\left( {D - Q} \right)\frac{1}{F} = \frac{1}{2} - \frac{1}{4} = 0.25 = 25\,cm\)
Option (2) is correct

Explanation:

Rays starting from \(O\) will suffer single refraction from spherical surface \(A P B\). Therefore, applying

\(\begin{aligned}& \dfrac{\mu_{2}}{v}-\dfrac{\mu_{1}}{u}=\dfrac{\mu_{2}-\mu_{1}}{R} ; \dfrac{1.0}{v}-\dfrac{2.0}{-15}=\dfrac{1.0-2.0}{-10} \\& =\dfrac{1}{v}=\dfrac{1}{10}-\dfrac{1}{7.5} \text { or } v=-30 {~cm}\end{aligned}\)
Therefore, image of \(O\) will be formed at 30 \(cm\) to the right of \(P\). Note that image will be virtual. There will be no effect of the surface \(C E D\).

Explanation:

Lenses have varying refractive indices for different wavelengths of light, with the refractive index decreasing as the wavelength increases. So, red light undergoes the least bending, while violet light undergoes the most.
For thickness of lews, there will be additional term added,

\(\frac{1}{f} = (\mu - 1)\left[ {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}} + \frac{{(\mu - 1)t}}{{\mu {R_1}{R_2}}}} \right]\)
\({f_{thick{\rm{ }}}} < {f_{thin{\rm{ }}}}\)
'Chromatic' means, we shall take 'colors' into account.
We know \(\lambda_{\text {red }}>\lambda_{\text {violet }}\)
\(\mu=\dfrac{c}{v}=\dfrac{c}{f \lambda} \Rightarrow \mu_{\text {red }} < \mu_{\text {violet }}\)
\(\Rightarrow f_{\text {red }}>f_{\text {violet }}\)
As they converge at different points, it is called 'aberration'
\(\Rightarrow\) Thick lens shows more aberration
\(\Rightarrow\) Assertion is true and reason is false.
So, correct option is (3).

Explanation:

Explanation:

Denote \(v_{O L}=\) Speed of object.
\(v_{I L}=\) Speed of image.
\(\Rightarrow v_{i L}=\left(\dfrac{v^{2}}{u^{2}}\right) v_{O L}\)
\(\Rightarrow 1=\dfrac{v^{2}}{u^{2}} \times 1\)
\(\Rightarrow v^{2}=u^{2}\)
We know, for lens:
\(\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\)
\(\Rightarrow v=\dfrac{u f}{u-f}\)
\(\Rightarrow\left(\dfrac{u f}{u-f}\right)^{2}=u^{2}\) [using equation (1)]
\( \Rightarrow u = 2f = 20\;cm\).
Correct option is (1).

Explanation:

\(\left( {A - R} \right)F = \frac{1}{P} = \frac{1}{1} = + 100\,cm\)
\(\left( {B - P} \right)\frac{1}{F} = \frac{1}{{0.5}} + \frac{1}{{0.5}} = + 400\,cm\)
\({\left( {C - S} \right)F = - \frac{1}{{2.5}} = - 0.4 = - 40\;cm}\)
\(\left( {D - Q} \right)\frac{1}{F} = \frac{1}{2} - \frac{1}{4} = 0.25 = 25\,cm\)
Option (2) is correct

Explanation:

Rays starting from \(O\) will suffer single refraction from spherical surface \(A P B\). Therefore, applying

\(\begin{aligned}& \dfrac{\mu_{2}}{v}-\dfrac{\mu_{1}}{u}=\dfrac{\mu_{2}-\mu_{1}}{R} ; \dfrac{1.0}{v}-\dfrac{2.0}{-15}=\dfrac{1.0-2.0}{-10} \\& =\dfrac{1}{v}=\dfrac{1}{10}-\dfrac{1}{7.5} \text { or } v=-30 {~cm}\end{aligned}\)
Therefore, image of \(O\) will be formed at 30 \(cm\) to the right of \(P\). Note that image will be virtual. There will be no effect of the surface \(C E D\).

Explanation:

Lenses have varying refractive indices for different wavelengths of light, with the refractive index decreasing as the wavelength increases. So, red light undergoes the least bending, while violet light undergoes the most.
For thickness of lews, there will be additional term added,

\(\frac{1}{f} = (\mu - 1)\left[ {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}} + \frac{{(\mu - 1)t}}{{\mu {R_1}{R_2}}}} \right]\)
\({f_{thick{\rm{ }}}} < {f_{thin{\rm{ }}}}\)
'Chromatic' means, we shall take 'colors' into account.
We know \(\lambda_{\text {red }}>\lambda_{\text {violet }}\)
\(\mu=\dfrac{c}{v}=\dfrac{c}{f \lambda} \Rightarrow \mu_{\text {red }} < \mu_{\text {violet }}\)
\(\Rightarrow f_{\text {red }}>f_{\text {violet }}\)
As they converge at different points, it is called 'aberration'
\(\Rightarrow\) Thick lens shows more aberration
\(\Rightarrow\) Assertion is true and reason is false.
So, correct option is (3).

Explanation:

Explanation:

Denote \(v_{O L}=\) Speed of object.
\(v_{I L}=\) Speed of image.
\(\Rightarrow v_{i L}=\left(\dfrac{v^{2}}{u^{2}}\right) v_{O L}\)
\(\Rightarrow 1=\dfrac{v^{2}}{u^{2}} \times 1\)
\(\Rightarrow v^{2}=u^{2}\)
We know, for lens:
\(\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\)
\(\Rightarrow v=\dfrac{u f}{u-f}\)
\(\Rightarrow\left(\dfrac{u f}{u-f}\right)^{2}=u^{2}\) [using equation (1)]
\( \Rightarrow u = 2f = 20\;cm\).
Correct option is (1).

Explanation:

\(\left( {A - R} \right)F = \frac{1}{P} = \frac{1}{1} = + 100\,cm\)
\(\left( {B - P} \right)\frac{1}{F} = \frac{1}{{0.5}} + \frac{1}{{0.5}} = + 400\,cm\)
\({\left( {C - S} \right)F = - \frac{1}{{2.5}} = - 0.4 = - 40\;cm}\)
\(\left( {D - Q} \right)\frac{1}{F} = \frac{1}{2} - \frac{1}{4} = 0.25 = 25\,cm\)
Option (2) is correct