364769
The position of image formed by the combination of lenses is
1 \(15\,cm\) (left of second lens)
2 \(30\,cm\) (right of third lens)
3 \(15\,cm\) (right of second lens)
4 \(30\,cm\) (left of third lens)
Explanation:
For \({1^{{\text{st}}}}\) Lens, \(u = - 30\;cm,f = 10\;cm\) \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \Rightarrow \frac{1}{{10}} = \frac{1}{v} - \frac{1}{{( - 30)}}\) \( \Rightarrow v = + 15\,cm\) For \({\text{I}}{{\text{I}}^{{\text{nd }}}}\) lens, the object is at its focus, so image is formed at infinity. For III \({ }^{\text {rd }}\) lens, objects at infinity. So final image is formed at its focus \(i.\,e.\) at a distance \(30\,cm\) from the right side of \(\mathrm{III}^{\mathrm{rd}}\) lens.
JEE - 2024
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364770
A real image is formed by a convex lens, when it is cemented with a concave lens, again the real image is formed. The real image
1 Shifts towards the lens system
2 Shifts away from the lens system
3 Remains in its original position
4 Shifts to infinity
Explanation:
Let \({f_1}\& {f_2}\)are the focal lengths of convex and concave lens. When these two lens are combined then \(\frac{1}{F} = \frac{1}{{{f_1}}} - \frac{1}{{{f_2}}} = \frac{{{f_2} - {f_1}}}{{{f_1}{f_2}}}\) Here \({f_2} > {f_1}\) since \(F\) is given as (+) ve \(F = \frac{{{f_1}{f_2}}}{{{f_2} - {f_1}}} > {f_1}\) i.e., the image shifts away from the lens
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364771
A thin convex lens of foal length \(10\,cm\) is placed in contact with the concave lens of same material and of same focal length. The focal length of combination will be:
1 Zero
2 Infinity
3 \(10\,cm\)
4 \(20\,cm\)
Explanation:
According to the relation \(\begin{aligned}& \dfrac{1}{f}=\dfrac{1}{f_{1}}+\dfrac{1}{f_{2}} \\& f=\dfrac{f_{1} f_{2}}{f_{1}+f_{2}}=\dfrac{10 \times(-10)}{10+(-10)}=\dfrac{-100}{0}=\infty .\end{aligned}\)
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364772
In the figure shown here, what is the equivalent focal length of the combination of lenses (Assume that all layers are thin)?
364769
The position of image formed by the combination of lenses is
1 \(15\,cm\) (left of second lens)
2 \(30\,cm\) (right of third lens)
3 \(15\,cm\) (right of second lens)
4 \(30\,cm\) (left of third lens)
Explanation:
For \({1^{{\text{st}}}}\) Lens, \(u = - 30\;cm,f = 10\;cm\) \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \Rightarrow \frac{1}{{10}} = \frac{1}{v} - \frac{1}{{( - 30)}}\) \( \Rightarrow v = + 15\,cm\) For \({\text{I}}{{\text{I}}^{{\text{nd }}}}\) lens, the object is at its focus, so image is formed at infinity. For III \({ }^{\text {rd }}\) lens, objects at infinity. So final image is formed at its focus \(i.\,e.\) at a distance \(30\,cm\) from the right side of \(\mathrm{III}^{\mathrm{rd}}\) lens.
JEE - 2024
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364770
A real image is formed by a convex lens, when it is cemented with a concave lens, again the real image is formed. The real image
1 Shifts towards the lens system
2 Shifts away from the lens system
3 Remains in its original position
4 Shifts to infinity
Explanation:
Let \({f_1}\& {f_2}\)are the focal lengths of convex and concave lens. When these two lens are combined then \(\frac{1}{F} = \frac{1}{{{f_1}}} - \frac{1}{{{f_2}}} = \frac{{{f_2} - {f_1}}}{{{f_1}{f_2}}}\) Here \({f_2} > {f_1}\) since \(F\) is given as (+) ve \(F = \frac{{{f_1}{f_2}}}{{{f_2} - {f_1}}} > {f_1}\) i.e., the image shifts away from the lens
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364771
A thin convex lens of foal length \(10\,cm\) is placed in contact with the concave lens of same material and of same focal length. The focal length of combination will be:
1 Zero
2 Infinity
3 \(10\,cm\)
4 \(20\,cm\)
Explanation:
According to the relation \(\begin{aligned}& \dfrac{1}{f}=\dfrac{1}{f_{1}}+\dfrac{1}{f_{2}} \\& f=\dfrac{f_{1} f_{2}}{f_{1}+f_{2}}=\dfrac{10 \times(-10)}{10+(-10)}=\dfrac{-100}{0}=\infty .\end{aligned}\)
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364772
In the figure shown here, what is the equivalent focal length of the combination of lenses (Assume that all layers are thin)?
364769
The position of image formed by the combination of lenses is
1 \(15\,cm\) (left of second lens)
2 \(30\,cm\) (right of third lens)
3 \(15\,cm\) (right of second lens)
4 \(30\,cm\) (left of third lens)
Explanation:
For \({1^{{\text{st}}}}\) Lens, \(u = - 30\;cm,f = 10\;cm\) \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \Rightarrow \frac{1}{{10}} = \frac{1}{v} - \frac{1}{{( - 30)}}\) \( \Rightarrow v = + 15\,cm\) For \({\text{I}}{{\text{I}}^{{\text{nd }}}}\) lens, the object is at its focus, so image is formed at infinity. For III \({ }^{\text {rd }}\) lens, objects at infinity. So final image is formed at its focus \(i.\,e.\) at a distance \(30\,cm\) from the right side of \(\mathrm{III}^{\mathrm{rd}}\) lens.
JEE - 2024
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364770
A real image is formed by a convex lens, when it is cemented with a concave lens, again the real image is formed. The real image
1 Shifts towards the lens system
2 Shifts away from the lens system
3 Remains in its original position
4 Shifts to infinity
Explanation:
Let \({f_1}\& {f_2}\)are the focal lengths of convex and concave lens. When these two lens are combined then \(\frac{1}{F} = \frac{1}{{{f_1}}} - \frac{1}{{{f_2}}} = \frac{{{f_2} - {f_1}}}{{{f_1}{f_2}}}\) Here \({f_2} > {f_1}\) since \(F\) is given as (+) ve \(F = \frac{{{f_1}{f_2}}}{{{f_2} - {f_1}}} > {f_1}\) i.e., the image shifts away from the lens
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364771
A thin convex lens of foal length \(10\,cm\) is placed in contact with the concave lens of same material and of same focal length. The focal length of combination will be:
1 Zero
2 Infinity
3 \(10\,cm\)
4 \(20\,cm\)
Explanation:
According to the relation \(\begin{aligned}& \dfrac{1}{f}=\dfrac{1}{f_{1}}+\dfrac{1}{f_{2}} \\& f=\dfrac{f_{1} f_{2}}{f_{1}+f_{2}}=\dfrac{10 \times(-10)}{10+(-10)}=\dfrac{-100}{0}=\infty .\end{aligned}\)
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364772
In the figure shown here, what is the equivalent focal length of the combination of lenses (Assume that all layers are thin)?
364769
The position of image formed by the combination of lenses is
1 \(15\,cm\) (left of second lens)
2 \(30\,cm\) (right of third lens)
3 \(15\,cm\) (right of second lens)
4 \(30\,cm\) (left of third lens)
Explanation:
For \({1^{{\text{st}}}}\) Lens, \(u = - 30\;cm,f = 10\;cm\) \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \Rightarrow \frac{1}{{10}} = \frac{1}{v} - \frac{1}{{( - 30)}}\) \( \Rightarrow v = + 15\,cm\) For \({\text{I}}{{\text{I}}^{{\text{nd }}}}\) lens, the object is at its focus, so image is formed at infinity. For III \({ }^{\text {rd }}\) lens, objects at infinity. So final image is formed at its focus \(i.\,e.\) at a distance \(30\,cm\) from the right side of \(\mathrm{III}^{\mathrm{rd}}\) lens.
JEE - 2024
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364770
A real image is formed by a convex lens, when it is cemented with a concave lens, again the real image is formed. The real image
1 Shifts towards the lens system
2 Shifts away from the lens system
3 Remains in its original position
4 Shifts to infinity
Explanation:
Let \({f_1}\& {f_2}\)are the focal lengths of convex and concave lens. When these two lens are combined then \(\frac{1}{F} = \frac{1}{{{f_1}}} - \frac{1}{{{f_2}}} = \frac{{{f_2} - {f_1}}}{{{f_1}{f_2}}}\) Here \({f_2} > {f_1}\) since \(F\) is given as (+) ve \(F = \frac{{{f_1}{f_2}}}{{{f_2} - {f_1}}} > {f_1}\) i.e., the image shifts away from the lens
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364771
A thin convex lens of foal length \(10\,cm\) is placed in contact with the concave lens of same material and of same focal length. The focal length of combination will be:
1 Zero
2 Infinity
3 \(10\,cm\)
4 \(20\,cm\)
Explanation:
According to the relation \(\begin{aligned}& \dfrac{1}{f}=\dfrac{1}{f_{1}}+\dfrac{1}{f_{2}} \\& f=\dfrac{f_{1} f_{2}}{f_{1}+f_{2}}=\dfrac{10 \times(-10)}{10+(-10)}=\dfrac{-100}{0}=\infty .\end{aligned}\)
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364772
In the figure shown here, what is the equivalent focal length of the combination of lenses (Assume that all layers are thin)?
