364825
Assertion : The focal length of lens does not change when red light is replaced by blue light. Reason : The focal length of lens depend on colour of light used.
1 Both Assertion and Reason are true and Reason is correct explanation of the Assertion.
2 Both Assertion and Reason are true but Reason is not the correct explanation of the Assertion.
3 Assertion is true but Reason is false.
4 Assertion is false but Reason is true.
Explanation:
Focal length of the lens depend upon it’s refractive index as \(\frac{1}{f} \propto (\mu - 1).\) Since \({\mu _b} < {\mu _r}\) so \({f_b} > {f_r}\) Therefore, the focal length of a lens decreases when red light is replaced by blue light. So option (4) is correct.
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364826
If the focal length of the convex lens is \(20\,cm \), what is the distance of the image from the lens in the following figure?
364827
The equi - convex lens has a focal length \('f'\). If the lens is cut along the line perpendicular to the principal axis and passing through the pole, what will be the focal length of any half part?
1 \(\frac{f}{2}\)
2 \(2f\)
3 \(\frac{{3f}}{2}\)
4 \(f\)
Explanation:
The lens Maker’s formula is given as \(\frac{1}{f} = \left( {{\mu _{med}} - 1} \right)\left( {\frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}} \right)\) Where, \(f\) = focal length of lens, \({R_1}\) = radius of first curved part and \({R_2}\)= radius of second curved part As, for equiconvex lens \({R_1} = {R_2} = R\left( {say} \right)\) So, \(\frac{1}{f} = \left( {{\mu _{real}} - 1} \right)\frac{2}{R}\;\;\;\;\;\;(1)\) Now, if lens is cut along the line perpendicular to the principal axis as shown in the figure The new cut part of the lens \({R_1} = R\) and \({R_2} = \infty \)
Again by using the lens maker’s formula, focal length of the new part of the lens \(\frac{1}{{f'}} = \left( {{\mu _{real}} - 1} \right)\left[ {\frac{1}{R} - \left( { - \frac{1}{\infty }} \right)} \right]\) \( \Rightarrow \frac{1}{{f'}} = \left( {{\mu _{real}} - 1} \right)\left[ {\frac{1}{R}} \right]\;\;\;\;\;\;(2)\) So, from the eq.(1) and eq.(2) , we get \({\rm{f}}' = 2f\)
MHTCET - 2019
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364828
Optic axis of a thin equiconvex lens is the\({\rm{ x}}\)-axis. The co-ordinates of a point object and its image are \(( - 40cm,{\rm{ }}1cm)\) and \((50cm,{\rm{ }} - 2cm)\) respectively. Lens is located at :-
1 \(x = - 30\;cm\)
2 \(x = + 20\;cm\)
3 \(x = 0\)
4 \(x = - 10\;cm\)
Explanation:
The object and image are shown in the figure From the figure \(m = \frac{{h'}}{h} = - 2 = \frac{{\left| v \right|}}{{ - \left| u \right|}}\) \( \Rightarrow \left| v \right| = 2\left| u \right|\) \({\left| v \right| + \left| u \right| = 90cm \Rightarrow \left| u \right| = 30\;cm}\) So the lens should be at \( - 10\;cm\)
364825
Assertion : The focal length of lens does not change when red light is replaced by blue light. Reason : The focal length of lens depend on colour of light used.
1 Both Assertion and Reason are true and Reason is correct explanation of the Assertion.
2 Both Assertion and Reason are true but Reason is not the correct explanation of the Assertion.
3 Assertion is true but Reason is false.
4 Assertion is false but Reason is true.
Explanation:
Focal length of the lens depend upon it’s refractive index as \(\frac{1}{f} \propto (\mu - 1).\) Since \({\mu _b} < {\mu _r}\) so \({f_b} > {f_r}\) Therefore, the focal length of a lens decreases when red light is replaced by blue light. So option (4) is correct.
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364826
If the focal length of the convex lens is \(20\,cm \), what is the distance of the image from the lens in the following figure?
364827
The equi - convex lens has a focal length \('f'\). If the lens is cut along the line perpendicular to the principal axis and passing through the pole, what will be the focal length of any half part?
1 \(\frac{f}{2}\)
2 \(2f\)
3 \(\frac{{3f}}{2}\)
4 \(f\)
Explanation:
The lens Maker’s formula is given as \(\frac{1}{f} = \left( {{\mu _{med}} - 1} \right)\left( {\frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}} \right)\) Where, \(f\) = focal length of lens, \({R_1}\) = radius of first curved part and \({R_2}\)= radius of second curved part As, for equiconvex lens \({R_1} = {R_2} = R\left( {say} \right)\) So, \(\frac{1}{f} = \left( {{\mu _{real}} - 1} \right)\frac{2}{R}\;\;\;\;\;\;(1)\) Now, if lens is cut along the line perpendicular to the principal axis as shown in the figure The new cut part of the lens \({R_1} = R\) and \({R_2} = \infty \)
Again by using the lens maker’s formula, focal length of the new part of the lens \(\frac{1}{{f'}} = \left( {{\mu _{real}} - 1} \right)\left[ {\frac{1}{R} - \left( { - \frac{1}{\infty }} \right)} \right]\) \( \Rightarrow \frac{1}{{f'}} = \left( {{\mu _{real}} - 1} \right)\left[ {\frac{1}{R}} \right]\;\;\;\;\;\;(2)\) So, from the eq.(1) and eq.(2) , we get \({\rm{f}}' = 2f\)
MHTCET - 2019
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364828
Optic axis of a thin equiconvex lens is the\({\rm{ x}}\)-axis. The co-ordinates of a point object and its image are \(( - 40cm,{\rm{ }}1cm)\) and \((50cm,{\rm{ }} - 2cm)\) respectively. Lens is located at :-
1 \(x = - 30\;cm\)
2 \(x = + 20\;cm\)
3 \(x = 0\)
4 \(x = - 10\;cm\)
Explanation:
The object and image are shown in the figure From the figure \(m = \frac{{h'}}{h} = - 2 = \frac{{\left| v \right|}}{{ - \left| u \right|}}\) \( \Rightarrow \left| v \right| = 2\left| u \right|\) \({\left| v \right| + \left| u \right| = 90cm \Rightarrow \left| u \right| = 30\;cm}\) So the lens should be at \( - 10\;cm\)
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PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364825
Assertion : The focal length of lens does not change when red light is replaced by blue light. Reason : The focal length of lens depend on colour of light used.
1 Both Assertion and Reason are true and Reason is correct explanation of the Assertion.
2 Both Assertion and Reason are true but Reason is not the correct explanation of the Assertion.
3 Assertion is true but Reason is false.
4 Assertion is false but Reason is true.
Explanation:
Focal length of the lens depend upon it’s refractive index as \(\frac{1}{f} \propto (\mu - 1).\) Since \({\mu _b} < {\mu _r}\) so \({f_b} > {f_r}\) Therefore, the focal length of a lens decreases when red light is replaced by blue light. So option (4) is correct.
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364826
If the focal length of the convex lens is \(20\,cm \), what is the distance of the image from the lens in the following figure?
364827
The equi - convex lens has a focal length \('f'\). If the lens is cut along the line perpendicular to the principal axis and passing through the pole, what will be the focal length of any half part?
1 \(\frac{f}{2}\)
2 \(2f\)
3 \(\frac{{3f}}{2}\)
4 \(f\)
Explanation:
The lens Maker’s formula is given as \(\frac{1}{f} = \left( {{\mu _{med}} - 1} \right)\left( {\frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}} \right)\) Where, \(f\) = focal length of lens, \({R_1}\) = radius of first curved part and \({R_2}\)= radius of second curved part As, for equiconvex lens \({R_1} = {R_2} = R\left( {say} \right)\) So, \(\frac{1}{f} = \left( {{\mu _{real}} - 1} \right)\frac{2}{R}\;\;\;\;\;\;(1)\) Now, if lens is cut along the line perpendicular to the principal axis as shown in the figure The new cut part of the lens \({R_1} = R\) and \({R_2} = \infty \)
Again by using the lens maker’s formula, focal length of the new part of the lens \(\frac{1}{{f'}} = \left( {{\mu _{real}} - 1} \right)\left[ {\frac{1}{R} - \left( { - \frac{1}{\infty }} \right)} \right]\) \( \Rightarrow \frac{1}{{f'}} = \left( {{\mu _{real}} - 1} \right)\left[ {\frac{1}{R}} \right]\;\;\;\;\;\;(2)\) So, from the eq.(1) and eq.(2) , we get \({\rm{f}}' = 2f\)
MHTCET - 2019
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364828
Optic axis of a thin equiconvex lens is the\({\rm{ x}}\)-axis. The co-ordinates of a point object and its image are \(( - 40cm,{\rm{ }}1cm)\) and \((50cm,{\rm{ }} - 2cm)\) respectively. Lens is located at :-
1 \(x = - 30\;cm\)
2 \(x = + 20\;cm\)
3 \(x = 0\)
4 \(x = - 10\;cm\)
Explanation:
The object and image are shown in the figure From the figure \(m = \frac{{h'}}{h} = - 2 = \frac{{\left| v \right|}}{{ - \left| u \right|}}\) \( \Rightarrow \left| v \right| = 2\left| u \right|\) \({\left| v \right| + \left| u \right| = 90cm \Rightarrow \left| u \right| = 30\;cm}\) So the lens should be at \( - 10\;cm\)
364825
Assertion : The focal length of lens does not change when red light is replaced by blue light. Reason : The focal length of lens depend on colour of light used.
1 Both Assertion and Reason are true and Reason is correct explanation of the Assertion.
2 Both Assertion and Reason are true but Reason is not the correct explanation of the Assertion.
3 Assertion is true but Reason is false.
4 Assertion is false but Reason is true.
Explanation:
Focal length of the lens depend upon it’s refractive index as \(\frac{1}{f} \propto (\mu - 1).\) Since \({\mu _b} < {\mu _r}\) so \({f_b} > {f_r}\) Therefore, the focal length of a lens decreases when red light is replaced by blue light. So option (4) is correct.
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364826
If the focal length of the convex lens is \(20\,cm \), what is the distance of the image from the lens in the following figure?
364827
The equi - convex lens has a focal length \('f'\). If the lens is cut along the line perpendicular to the principal axis and passing through the pole, what will be the focal length of any half part?
1 \(\frac{f}{2}\)
2 \(2f\)
3 \(\frac{{3f}}{2}\)
4 \(f\)
Explanation:
The lens Maker’s formula is given as \(\frac{1}{f} = \left( {{\mu _{med}} - 1} \right)\left( {\frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}} \right)\) Where, \(f\) = focal length of lens, \({R_1}\) = radius of first curved part and \({R_2}\)= radius of second curved part As, for equiconvex lens \({R_1} = {R_2} = R\left( {say} \right)\) So, \(\frac{1}{f} = \left( {{\mu _{real}} - 1} \right)\frac{2}{R}\;\;\;\;\;\;(1)\) Now, if lens is cut along the line perpendicular to the principal axis as shown in the figure The new cut part of the lens \({R_1} = R\) and \({R_2} = \infty \)
Again by using the lens maker’s formula, focal length of the new part of the lens \(\frac{1}{{f'}} = \left( {{\mu _{real}} - 1} \right)\left[ {\frac{1}{R} - \left( { - \frac{1}{\infty }} \right)} \right]\) \( \Rightarrow \frac{1}{{f'}} = \left( {{\mu _{real}} - 1} \right)\left[ {\frac{1}{R}} \right]\;\;\;\;\;\;(2)\) So, from the eq.(1) and eq.(2) , we get \({\rm{f}}' = 2f\)
MHTCET - 2019
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364828
Optic axis of a thin equiconvex lens is the\({\rm{ x}}\)-axis. The co-ordinates of a point object and its image are \(( - 40cm,{\rm{ }}1cm)\) and \((50cm,{\rm{ }} - 2cm)\) respectively. Lens is located at :-
1 \(x = - 30\;cm\)
2 \(x = + 20\;cm\)
3 \(x = 0\)
4 \(x = - 10\;cm\)
Explanation:
The object and image are shown in the figure From the figure \(m = \frac{{h'}}{h} = - 2 = \frac{{\left| v \right|}}{{ - \left| u \right|}}\) \( \Rightarrow \left| v \right| = 2\left| u \right|\) \({\left| v \right| + \left| u \right| = 90cm \Rightarrow \left| u \right| = 30\;cm}\) So the lens should be at \( - 10\;cm\)
