NEET Test Series from KOTA - 10 Papers In MS WORD
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Ray Optics
282411
An object is placed at a distance of \(20 \mathrm{~cm}\) from a lens as shown in the figure. Find the focal length of lens if the magnification is \(\mathbf{- 0 . 5}\).
282412
A person wears normal spectacles in which the distance of glasses and eyes is approximately 2 \(\mathrm{cm}\), then power required is \(-5 \mathrm{D}\). If wears contact lens, then the required power is:
1 \(-5.2 \mathrm{D}\)
2 \(-4.54 \mathrm{D}\)
3 \(+5.2 \mathrm{D}\)
4 \(+4.7 \mathrm{D}\)
Explanation:
B: Given,
Power, \(\mathrm{P}=-5 \mathrm{D}\)
Contact lens is more effective, so it's required less power-
For glass \(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}\)
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\frac{1}{(-\mathrm{x})}-\frac{1}{\infty} \\
\mathrm{f}=-\mathrm{x} \mathrm{cm}
\end{aligned}\)
Power, \(\mathrm{P}=\frac{1}{\mathrm{f}}=\frac{100}{-\mathrm{x}}=-5\)
\(\Rightarrow \mathrm{x}=20 \mathrm{~cm}\)
If he use contact lens -
Again,
\(\mathrm{V}=-\infty, \mathrm{u}=-22 \mathrm{~cm}\)
\(\begin{aligned}
\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}^{\prime}} \\
\frac{1}{-\infty}-\frac{1}{22}=\frac{1}{\mathrm{f}^{\prime}} \\
\mathrm{f}^{\prime}=-22 \mathrm{~cm}
\end{aligned}\)
Power, \(\mathrm{P}^{\prime}=\frac{1}{\mathrm{f}^{\prime}}=\frac{100}{-22}\)
\(\mathrm{P}^{\prime}=-4.54 \mathrm{D}\)
AIIMS-25.05.2019(M) Shift-1
Ray Optics
282413
Assertion: We cannot produce a real image by plane or convex mirrors under any circumstances.
Reason: The focal length of a convex mirror is always taken negative.
1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
D: Focal length of convex mirror is taken positive.
AIIMS-25.05.2019(E) Shift-2
Ray Optics
282414
The focal length of a thin lens made from the material of refractive index 1.5 is \(15 \mathrm{~cm}\). When it is placed in a liquid of refractive index \(\frac{4}{3}\), its focal length will be cm.
1 80.31
2 50
3 78.23
4 60
Explanation:
D: Given,
Refractive index of glass \(\left(\mu_{\mathrm{g}}\right)=1.5\),
Focal length of lens \((\mathrm{f})=15 \mathrm{~cm}\),
Refractive index of liquid \(\left(\mu_{\mathrm{w}}\right)=4 / 3\)
We know that,
\(\begin{gathered}
\frac{\mathrm{f}^{\prime}}{\mathrm{f}}=\frac{\left(\mu_{\mathrm{g}}-1\right)}{\left(\mu_{\mathrm{g}} / \mu_{\mathrm{w}}-1\right)} \\
\mathrm{f}^{\prime}=15 \times \frac{(1.5-1)}{\left(\frac{1.5}{4 / 3}-1\right)} \\
\mathrm{f}^{\prime}=\frac{15 \times 0.5 \times 4}{0.5} \\
\mathrm{f}^{\prime}=60 \mathrm{~cm}
\end{gathered}\)
282411
An object is placed at a distance of \(20 \mathrm{~cm}\) from a lens as shown in the figure. Find the focal length of lens if the magnification is \(\mathbf{- 0 . 5}\).
282412
A person wears normal spectacles in which the distance of glasses and eyes is approximately 2 \(\mathrm{cm}\), then power required is \(-5 \mathrm{D}\). If wears contact lens, then the required power is:
1 \(-5.2 \mathrm{D}\)
2 \(-4.54 \mathrm{D}\)
3 \(+5.2 \mathrm{D}\)
4 \(+4.7 \mathrm{D}\)
Explanation:
B: Given,
Power, \(\mathrm{P}=-5 \mathrm{D}\)
Contact lens is more effective, so it's required less power-
For glass \(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}\)
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\frac{1}{(-\mathrm{x})}-\frac{1}{\infty} \\
\mathrm{f}=-\mathrm{x} \mathrm{cm}
\end{aligned}\)
Power, \(\mathrm{P}=\frac{1}{\mathrm{f}}=\frac{100}{-\mathrm{x}}=-5\)
\(\Rightarrow \mathrm{x}=20 \mathrm{~cm}\)
If he use contact lens -
Again,
\(\mathrm{V}=-\infty, \mathrm{u}=-22 \mathrm{~cm}\)
\(\begin{aligned}
\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}^{\prime}} \\
\frac{1}{-\infty}-\frac{1}{22}=\frac{1}{\mathrm{f}^{\prime}} \\
\mathrm{f}^{\prime}=-22 \mathrm{~cm}
\end{aligned}\)
Power, \(\mathrm{P}^{\prime}=\frac{1}{\mathrm{f}^{\prime}}=\frac{100}{-22}\)
\(\mathrm{P}^{\prime}=-4.54 \mathrm{D}\)
AIIMS-25.05.2019(M) Shift-1
Ray Optics
282413
Assertion: We cannot produce a real image by plane or convex mirrors under any circumstances.
Reason: The focal length of a convex mirror is always taken negative.
1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
D: Focal length of convex mirror is taken positive.
AIIMS-25.05.2019(E) Shift-2
Ray Optics
282414
The focal length of a thin lens made from the material of refractive index 1.5 is \(15 \mathrm{~cm}\). When it is placed in a liquid of refractive index \(\frac{4}{3}\), its focal length will be cm.
1 80.31
2 50
3 78.23
4 60
Explanation:
D: Given,
Refractive index of glass \(\left(\mu_{\mathrm{g}}\right)=1.5\),
Focal length of lens \((\mathrm{f})=15 \mathrm{~cm}\),
Refractive index of liquid \(\left(\mu_{\mathrm{w}}\right)=4 / 3\)
We know that,
\(\begin{gathered}
\frac{\mathrm{f}^{\prime}}{\mathrm{f}}=\frac{\left(\mu_{\mathrm{g}}-1\right)}{\left(\mu_{\mathrm{g}} / \mu_{\mathrm{w}}-1\right)} \\
\mathrm{f}^{\prime}=15 \times \frac{(1.5-1)}{\left(\frac{1.5}{4 / 3}-1\right)} \\
\mathrm{f}^{\prime}=\frac{15 \times 0.5 \times 4}{0.5} \\
\mathrm{f}^{\prime}=60 \mathrm{~cm}
\end{gathered}\)
282411
An object is placed at a distance of \(20 \mathrm{~cm}\) from a lens as shown in the figure. Find the focal length of lens if the magnification is \(\mathbf{- 0 . 5}\).
282412
A person wears normal spectacles in which the distance of glasses and eyes is approximately 2 \(\mathrm{cm}\), then power required is \(-5 \mathrm{D}\). If wears contact lens, then the required power is:
1 \(-5.2 \mathrm{D}\)
2 \(-4.54 \mathrm{D}\)
3 \(+5.2 \mathrm{D}\)
4 \(+4.7 \mathrm{D}\)
Explanation:
B: Given,
Power, \(\mathrm{P}=-5 \mathrm{D}\)
Contact lens is more effective, so it's required less power-
For glass \(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}\)
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\frac{1}{(-\mathrm{x})}-\frac{1}{\infty} \\
\mathrm{f}=-\mathrm{x} \mathrm{cm}
\end{aligned}\)
Power, \(\mathrm{P}=\frac{1}{\mathrm{f}}=\frac{100}{-\mathrm{x}}=-5\)
\(\Rightarrow \mathrm{x}=20 \mathrm{~cm}\)
If he use contact lens -
Again,
\(\mathrm{V}=-\infty, \mathrm{u}=-22 \mathrm{~cm}\)
\(\begin{aligned}
\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}^{\prime}} \\
\frac{1}{-\infty}-\frac{1}{22}=\frac{1}{\mathrm{f}^{\prime}} \\
\mathrm{f}^{\prime}=-22 \mathrm{~cm}
\end{aligned}\)
Power, \(\mathrm{P}^{\prime}=\frac{1}{\mathrm{f}^{\prime}}=\frac{100}{-22}\)
\(\mathrm{P}^{\prime}=-4.54 \mathrm{D}\)
AIIMS-25.05.2019(M) Shift-1
Ray Optics
282413
Assertion: We cannot produce a real image by plane or convex mirrors under any circumstances.
Reason: The focal length of a convex mirror is always taken negative.
1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
D: Focal length of convex mirror is taken positive.
AIIMS-25.05.2019(E) Shift-2
Ray Optics
282414
The focal length of a thin lens made from the material of refractive index 1.5 is \(15 \mathrm{~cm}\). When it is placed in a liquid of refractive index \(\frac{4}{3}\), its focal length will be cm.
1 80.31
2 50
3 78.23
4 60
Explanation:
D: Given,
Refractive index of glass \(\left(\mu_{\mathrm{g}}\right)=1.5\),
Focal length of lens \((\mathrm{f})=15 \mathrm{~cm}\),
Refractive index of liquid \(\left(\mu_{\mathrm{w}}\right)=4 / 3\)
We know that,
\(\begin{gathered}
\frac{\mathrm{f}^{\prime}}{\mathrm{f}}=\frac{\left(\mu_{\mathrm{g}}-1\right)}{\left(\mu_{\mathrm{g}} / \mu_{\mathrm{w}}-1\right)} \\
\mathrm{f}^{\prime}=15 \times \frac{(1.5-1)}{\left(\frac{1.5}{4 / 3}-1\right)} \\
\mathrm{f}^{\prime}=\frac{15 \times 0.5 \times 4}{0.5} \\
\mathrm{f}^{\prime}=60 \mathrm{~cm}
\end{gathered}\)
282411
An object is placed at a distance of \(20 \mathrm{~cm}\) from a lens as shown in the figure. Find the focal length of lens if the magnification is \(\mathbf{- 0 . 5}\).
282412
A person wears normal spectacles in which the distance of glasses and eyes is approximately 2 \(\mathrm{cm}\), then power required is \(-5 \mathrm{D}\). If wears contact lens, then the required power is:
1 \(-5.2 \mathrm{D}\)
2 \(-4.54 \mathrm{D}\)
3 \(+5.2 \mathrm{D}\)
4 \(+4.7 \mathrm{D}\)
Explanation:
B: Given,
Power, \(\mathrm{P}=-5 \mathrm{D}\)
Contact lens is more effective, so it's required less power-
For glass \(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}\)
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\frac{1}{(-\mathrm{x})}-\frac{1}{\infty} \\
\mathrm{f}=-\mathrm{x} \mathrm{cm}
\end{aligned}\)
Power, \(\mathrm{P}=\frac{1}{\mathrm{f}}=\frac{100}{-\mathrm{x}}=-5\)
\(\Rightarrow \mathrm{x}=20 \mathrm{~cm}\)
If he use contact lens -
Again,
\(\mathrm{V}=-\infty, \mathrm{u}=-22 \mathrm{~cm}\)
\(\begin{aligned}
\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}^{\prime}} \\
\frac{1}{-\infty}-\frac{1}{22}=\frac{1}{\mathrm{f}^{\prime}} \\
\mathrm{f}^{\prime}=-22 \mathrm{~cm}
\end{aligned}\)
Power, \(\mathrm{P}^{\prime}=\frac{1}{\mathrm{f}^{\prime}}=\frac{100}{-22}\)
\(\mathrm{P}^{\prime}=-4.54 \mathrm{D}\)
AIIMS-25.05.2019(M) Shift-1
Ray Optics
282413
Assertion: We cannot produce a real image by plane or convex mirrors under any circumstances.
Reason: The focal length of a convex mirror is always taken negative.
1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
D: Focal length of convex mirror is taken positive.
AIIMS-25.05.2019(E) Shift-2
Ray Optics
282414
The focal length of a thin lens made from the material of refractive index 1.5 is \(15 \mathrm{~cm}\). When it is placed in a liquid of refractive index \(\frac{4}{3}\), its focal length will be cm.
1 80.31
2 50
3 78.23
4 60
Explanation:
D: Given,
Refractive index of glass \(\left(\mu_{\mathrm{g}}\right)=1.5\),
Focal length of lens \((\mathrm{f})=15 \mathrm{~cm}\),
Refractive index of liquid \(\left(\mu_{\mathrm{w}}\right)=4 / 3\)
We know that,
\(\begin{gathered}
\frac{\mathrm{f}^{\prime}}{\mathrm{f}}=\frac{\left(\mu_{\mathrm{g}}-1\right)}{\left(\mu_{\mathrm{g}} / \mu_{\mathrm{w}}-1\right)} \\
\mathrm{f}^{\prime}=15 \times \frac{(1.5-1)}{\left(\frac{1.5}{4 / 3}-1\right)} \\
\mathrm{f}^{\prime}=\frac{15 \times 0.5 \times 4}{0.5} \\
\mathrm{f}^{\prime}=60 \mathrm{~cm}
\end{gathered}\)
