364953
A boat has green light of wavelength \(\lambda = 500\;nm\) on the mast. What wavelength would be measured and what colour would be observed for this light as seen by a driver submerged in water by the side of the boat? Given \({n_w} = \frac{4}{3}\)
1 Green of wavelength \(375\;nm\)
2 \(665\;nm\)
3 Green of waelength \(500\;nm\)
4 Blue of wavelength \(375\;nm\)
Explanation:
Here, \({n_w} = \frac{4}{3},{\mkern 1mu} {\mkern 1mu} {\lambda _a} = 500nm\) As \({n_w} = \frac{{{\lambda _a}}}{{{\lambda _w}}}\), [subscripts \(a\) and \(w\) for air and water respectively] \(\therefore {\lambda _w} = \frac{{500\;nm}}{{\left( {4/3} \right)}} = 375\;nm\) Colour of light is determined by its frequency and not by its wavelength. As frequency does not change on refraction, colour will not change and will remain green.
KCET - 2010
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364954
Total internal reflection takes place
1 when a ray moves from denser to rarer and incident angle is greater than critical angle
2 when a ray moves from rarer to denser and incident angle is less than critical angle
3 when a ray moves from rarer to denser and incident angle is equal to critical angle
4 None of the above
Explanation:
Total internal reflection takes place when ray goes from denser to rarer medium and incident angle is greaterthan critical angle.
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364955
Assertion : Endoscopy involves use of optical fibres to study internal organs. Reason : Optical fibres are based on phenomenon of total internal reflection.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Endoscopy involves the use of optical fibers to study internal organs. This is correct. Endoscopy is a medical procedure that utilizes thin, flexible optical fibers with a light source and a camera to visualize and study internal organs or cavities. The Reason provided explains why optical fibers are used in endoscopy. It correctly states that optical fibers are based on the phenomenon of total internal reflection. This phenomenon allows light to be transmitted through the fibers by repeatedly undergoing total internal reflection, which enables the transmission of images and illumination in endoscopy procedures. The Reason provides a valid and accurate explanation for the Assertion, so the correct choice is (1).
AIIMS - 2009
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364956
A fish looking up through the water sees the outside world, contained in a circular horizon. If the refractive index of water is \(\dfrac{4}{3}\) and the fish is \(12\;cm\) below the water surface, the radius of this circle in \(cm\) is
1 \(36 \sqrt{7}\)
2 \(\dfrac{36}{\sqrt{7}}\)
3 \(36 \sqrt{5}\)
4 \(4 \sqrt{5}\)
Explanation:
The situtaiton is shown in figure \(\begin{aligned}& \sin \theta_{C}=\dfrac{1}{\mu} \Rightarrow \tan \theta_{C}=\dfrac{A B}{A O}\left(\Rightarrow \sin \theta_{c}=\dfrac{3}{4}\right) \\& \Rightarrow A B=O A \tan \theta_{C}\end{aligned}\) \(\begin{gathered}A B=\dfrac{O A}{\sqrt{\mu^{2}-1}}=\dfrac{12}{\sqrt{\left(\dfrac{4}{3}\right)^{2}-1}}=\dfrac{36}{\sqrt{7}} \\\quad\left[\therefore \tan \theta_{C}=\dfrac{1}{\sqrt{\mu^{2}-1}}\right]\end{gathered}\)
AIIMS - 2008
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364957
In optical fibres, the refractive index of the core is
1 greater than that of the cladding
2 equal to that of the cladding
3 smaller than that of the cladding
4 independent of that of the cladding
Explanation:
In optical fibres, the refractive index of the core is greater than that of the cladding. Thus, the core act as denser medium and cladding as rarer medium. For angle of incidence greater than critical angle, the ray forward through total internal reflection.
364953
A boat has green light of wavelength \(\lambda = 500\;nm\) on the mast. What wavelength would be measured and what colour would be observed for this light as seen by a driver submerged in water by the side of the boat? Given \({n_w} = \frac{4}{3}\)
1 Green of wavelength \(375\;nm\)
2 \(665\;nm\)
3 Green of waelength \(500\;nm\)
4 Blue of wavelength \(375\;nm\)
Explanation:
Here, \({n_w} = \frac{4}{3},{\mkern 1mu} {\mkern 1mu} {\lambda _a} = 500nm\) As \({n_w} = \frac{{{\lambda _a}}}{{{\lambda _w}}}\), [subscripts \(a\) and \(w\) for air and water respectively] \(\therefore {\lambda _w} = \frac{{500\;nm}}{{\left( {4/3} \right)}} = 375\;nm\) Colour of light is determined by its frequency and not by its wavelength. As frequency does not change on refraction, colour will not change and will remain green.
KCET - 2010
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364954
Total internal reflection takes place
1 when a ray moves from denser to rarer and incident angle is greater than critical angle
2 when a ray moves from rarer to denser and incident angle is less than critical angle
3 when a ray moves from rarer to denser and incident angle is equal to critical angle
4 None of the above
Explanation:
Total internal reflection takes place when ray goes from denser to rarer medium and incident angle is greaterthan critical angle.
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364955
Assertion : Endoscopy involves use of optical fibres to study internal organs. Reason : Optical fibres are based on phenomenon of total internal reflection.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Endoscopy involves the use of optical fibers to study internal organs. This is correct. Endoscopy is a medical procedure that utilizes thin, flexible optical fibers with a light source and a camera to visualize and study internal organs or cavities. The Reason provided explains why optical fibers are used in endoscopy. It correctly states that optical fibers are based on the phenomenon of total internal reflection. This phenomenon allows light to be transmitted through the fibers by repeatedly undergoing total internal reflection, which enables the transmission of images and illumination in endoscopy procedures. The Reason provides a valid and accurate explanation for the Assertion, so the correct choice is (1).
AIIMS - 2009
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364956
A fish looking up through the water sees the outside world, contained in a circular horizon. If the refractive index of water is \(\dfrac{4}{3}\) and the fish is \(12\;cm\) below the water surface, the radius of this circle in \(cm\) is
1 \(36 \sqrt{7}\)
2 \(\dfrac{36}{\sqrt{7}}\)
3 \(36 \sqrt{5}\)
4 \(4 \sqrt{5}\)
Explanation:
The situtaiton is shown in figure \(\begin{aligned}& \sin \theta_{C}=\dfrac{1}{\mu} \Rightarrow \tan \theta_{C}=\dfrac{A B}{A O}\left(\Rightarrow \sin \theta_{c}=\dfrac{3}{4}\right) \\& \Rightarrow A B=O A \tan \theta_{C}\end{aligned}\) \(\begin{gathered}A B=\dfrac{O A}{\sqrt{\mu^{2}-1}}=\dfrac{12}{\sqrt{\left(\dfrac{4}{3}\right)^{2}-1}}=\dfrac{36}{\sqrt{7}} \\\quad\left[\therefore \tan \theta_{C}=\dfrac{1}{\sqrt{\mu^{2}-1}}\right]\end{gathered}\)
AIIMS - 2008
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364957
In optical fibres, the refractive index of the core is
1 greater than that of the cladding
2 equal to that of the cladding
3 smaller than that of the cladding
4 independent of that of the cladding
Explanation:
In optical fibres, the refractive index of the core is greater than that of the cladding. Thus, the core act as denser medium and cladding as rarer medium. For angle of incidence greater than critical angle, the ray forward through total internal reflection.
364953
A boat has green light of wavelength \(\lambda = 500\;nm\) on the mast. What wavelength would be measured and what colour would be observed for this light as seen by a driver submerged in water by the side of the boat? Given \({n_w} = \frac{4}{3}\)
1 Green of wavelength \(375\;nm\)
2 \(665\;nm\)
3 Green of waelength \(500\;nm\)
4 Blue of wavelength \(375\;nm\)
Explanation:
Here, \({n_w} = \frac{4}{3},{\mkern 1mu} {\mkern 1mu} {\lambda _a} = 500nm\) As \({n_w} = \frac{{{\lambda _a}}}{{{\lambda _w}}}\), [subscripts \(a\) and \(w\) for air and water respectively] \(\therefore {\lambda _w} = \frac{{500\;nm}}{{\left( {4/3} \right)}} = 375\;nm\) Colour of light is determined by its frequency and not by its wavelength. As frequency does not change on refraction, colour will not change and will remain green.
KCET - 2010
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364954
Total internal reflection takes place
1 when a ray moves from denser to rarer and incident angle is greater than critical angle
2 when a ray moves from rarer to denser and incident angle is less than critical angle
3 when a ray moves from rarer to denser and incident angle is equal to critical angle
4 None of the above
Explanation:
Total internal reflection takes place when ray goes from denser to rarer medium and incident angle is greaterthan critical angle.
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364955
Assertion : Endoscopy involves use of optical fibres to study internal organs. Reason : Optical fibres are based on phenomenon of total internal reflection.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Endoscopy involves the use of optical fibers to study internal organs. This is correct. Endoscopy is a medical procedure that utilizes thin, flexible optical fibers with a light source and a camera to visualize and study internal organs or cavities. The Reason provided explains why optical fibers are used in endoscopy. It correctly states that optical fibers are based on the phenomenon of total internal reflection. This phenomenon allows light to be transmitted through the fibers by repeatedly undergoing total internal reflection, which enables the transmission of images and illumination in endoscopy procedures. The Reason provides a valid and accurate explanation for the Assertion, so the correct choice is (1).
AIIMS - 2009
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364956
A fish looking up through the water sees the outside world, contained in a circular horizon. If the refractive index of water is \(\dfrac{4}{3}\) and the fish is \(12\;cm\) below the water surface, the radius of this circle in \(cm\) is
1 \(36 \sqrt{7}\)
2 \(\dfrac{36}{\sqrt{7}}\)
3 \(36 \sqrt{5}\)
4 \(4 \sqrt{5}\)
Explanation:
The situtaiton is shown in figure \(\begin{aligned}& \sin \theta_{C}=\dfrac{1}{\mu} \Rightarrow \tan \theta_{C}=\dfrac{A B}{A O}\left(\Rightarrow \sin \theta_{c}=\dfrac{3}{4}\right) \\& \Rightarrow A B=O A \tan \theta_{C}\end{aligned}\) \(\begin{gathered}A B=\dfrac{O A}{\sqrt{\mu^{2}-1}}=\dfrac{12}{\sqrt{\left(\dfrac{4}{3}\right)^{2}-1}}=\dfrac{36}{\sqrt{7}} \\\quad\left[\therefore \tan \theta_{C}=\dfrac{1}{\sqrt{\mu^{2}-1}}\right]\end{gathered}\)
AIIMS - 2008
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364957
In optical fibres, the refractive index of the core is
1 greater than that of the cladding
2 equal to that of the cladding
3 smaller than that of the cladding
4 independent of that of the cladding
Explanation:
In optical fibres, the refractive index of the core is greater than that of the cladding. Thus, the core act as denser medium and cladding as rarer medium. For angle of incidence greater than critical angle, the ray forward through total internal reflection.
364953
A boat has green light of wavelength \(\lambda = 500\;nm\) on the mast. What wavelength would be measured and what colour would be observed for this light as seen by a driver submerged in water by the side of the boat? Given \({n_w} = \frac{4}{3}\)
1 Green of wavelength \(375\;nm\)
2 \(665\;nm\)
3 Green of waelength \(500\;nm\)
4 Blue of wavelength \(375\;nm\)
Explanation:
Here, \({n_w} = \frac{4}{3},{\mkern 1mu} {\mkern 1mu} {\lambda _a} = 500nm\) As \({n_w} = \frac{{{\lambda _a}}}{{{\lambda _w}}}\), [subscripts \(a\) and \(w\) for air and water respectively] \(\therefore {\lambda _w} = \frac{{500\;nm}}{{\left( {4/3} \right)}} = 375\;nm\) Colour of light is determined by its frequency and not by its wavelength. As frequency does not change on refraction, colour will not change and will remain green.
KCET - 2010
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364954
Total internal reflection takes place
1 when a ray moves from denser to rarer and incident angle is greater than critical angle
2 when a ray moves from rarer to denser and incident angle is less than critical angle
3 when a ray moves from rarer to denser and incident angle is equal to critical angle
4 None of the above
Explanation:
Total internal reflection takes place when ray goes from denser to rarer medium and incident angle is greaterthan critical angle.
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364955
Assertion : Endoscopy involves use of optical fibres to study internal organs. Reason : Optical fibres are based on phenomenon of total internal reflection.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Endoscopy involves the use of optical fibers to study internal organs. This is correct. Endoscopy is a medical procedure that utilizes thin, flexible optical fibers with a light source and a camera to visualize and study internal organs or cavities. The Reason provided explains why optical fibers are used in endoscopy. It correctly states that optical fibers are based on the phenomenon of total internal reflection. This phenomenon allows light to be transmitted through the fibers by repeatedly undergoing total internal reflection, which enables the transmission of images and illumination in endoscopy procedures. The Reason provides a valid and accurate explanation for the Assertion, so the correct choice is (1).
AIIMS - 2009
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364956
A fish looking up through the water sees the outside world, contained in a circular horizon. If the refractive index of water is \(\dfrac{4}{3}\) and the fish is \(12\;cm\) below the water surface, the radius of this circle in \(cm\) is
1 \(36 \sqrt{7}\)
2 \(\dfrac{36}{\sqrt{7}}\)
3 \(36 \sqrt{5}\)
4 \(4 \sqrt{5}\)
Explanation:
The situtaiton is shown in figure \(\begin{aligned}& \sin \theta_{C}=\dfrac{1}{\mu} \Rightarrow \tan \theta_{C}=\dfrac{A B}{A O}\left(\Rightarrow \sin \theta_{c}=\dfrac{3}{4}\right) \\& \Rightarrow A B=O A \tan \theta_{C}\end{aligned}\) \(\begin{gathered}A B=\dfrac{O A}{\sqrt{\mu^{2}-1}}=\dfrac{12}{\sqrt{\left(\dfrac{4}{3}\right)^{2}-1}}=\dfrac{36}{\sqrt{7}} \\\quad\left[\therefore \tan \theta_{C}=\dfrac{1}{\sqrt{\mu^{2}-1}}\right]\end{gathered}\)
AIIMS - 2008
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364957
In optical fibres, the refractive index of the core is
1 greater than that of the cladding
2 equal to that of the cladding
3 smaller than that of the cladding
4 independent of that of the cladding
Explanation:
In optical fibres, the refractive index of the core is greater than that of the cladding. Thus, the core act as denser medium and cladding as rarer medium. For angle of incidence greater than critical angle, the ray forward through total internal reflection.
364953
A boat has green light of wavelength \(\lambda = 500\;nm\) on the mast. What wavelength would be measured and what colour would be observed for this light as seen by a driver submerged in water by the side of the boat? Given \({n_w} = \frac{4}{3}\)
1 Green of wavelength \(375\;nm\)
2 \(665\;nm\)
3 Green of waelength \(500\;nm\)
4 Blue of wavelength \(375\;nm\)
Explanation:
Here, \({n_w} = \frac{4}{3},{\mkern 1mu} {\mkern 1mu} {\lambda _a} = 500nm\) As \({n_w} = \frac{{{\lambda _a}}}{{{\lambda _w}}}\), [subscripts \(a\) and \(w\) for air and water respectively] \(\therefore {\lambda _w} = \frac{{500\;nm}}{{\left( {4/3} \right)}} = 375\;nm\) Colour of light is determined by its frequency and not by its wavelength. As frequency does not change on refraction, colour will not change and will remain green.
KCET - 2010
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364954
Total internal reflection takes place
1 when a ray moves from denser to rarer and incident angle is greater than critical angle
2 when a ray moves from rarer to denser and incident angle is less than critical angle
3 when a ray moves from rarer to denser and incident angle is equal to critical angle
4 None of the above
Explanation:
Total internal reflection takes place when ray goes from denser to rarer medium and incident angle is greaterthan critical angle.
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364955
Assertion : Endoscopy involves use of optical fibres to study internal organs. Reason : Optical fibres are based on phenomenon of total internal reflection.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Endoscopy involves the use of optical fibers to study internal organs. This is correct. Endoscopy is a medical procedure that utilizes thin, flexible optical fibers with a light source and a camera to visualize and study internal organs or cavities. The Reason provided explains why optical fibers are used in endoscopy. It correctly states that optical fibers are based on the phenomenon of total internal reflection. This phenomenon allows light to be transmitted through the fibers by repeatedly undergoing total internal reflection, which enables the transmission of images and illumination in endoscopy procedures. The Reason provides a valid and accurate explanation for the Assertion, so the correct choice is (1).
AIIMS - 2009
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364956
A fish looking up through the water sees the outside world, contained in a circular horizon. If the refractive index of water is \(\dfrac{4}{3}\) and the fish is \(12\;cm\) below the water surface, the radius of this circle in \(cm\) is
1 \(36 \sqrt{7}\)
2 \(\dfrac{36}{\sqrt{7}}\)
3 \(36 \sqrt{5}\)
4 \(4 \sqrt{5}\)
Explanation:
The situtaiton is shown in figure \(\begin{aligned}& \sin \theta_{C}=\dfrac{1}{\mu} \Rightarrow \tan \theta_{C}=\dfrac{A B}{A O}\left(\Rightarrow \sin \theta_{c}=\dfrac{3}{4}\right) \\& \Rightarrow A B=O A \tan \theta_{C}\end{aligned}\) \(\begin{gathered}A B=\dfrac{O A}{\sqrt{\mu^{2}-1}}=\dfrac{12}{\sqrt{\left(\dfrac{4}{3}\right)^{2}-1}}=\dfrac{36}{\sqrt{7}} \\\quad\left[\therefore \tan \theta_{C}=\dfrac{1}{\sqrt{\mu^{2}-1}}\right]\end{gathered}\)
AIIMS - 2008
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364957
In optical fibres, the refractive index of the core is
1 greater than that of the cladding
2 equal to that of the cladding
3 smaller than that of the cladding
4 independent of that of the cladding
Explanation:
In optical fibres, the refractive index of the core is greater than that of the cladding. Thus, the core act as denser medium and cladding as rarer medium. For angle of incidence greater than critical angle, the ray forward through total internal reflection.
