JEE & NEET Physics Quiz: Ray Optics and Optical Instruments
Total Marks: 40
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1. A ray is incident at 30° angle on a plane mirror. What will be the deviation after reflection from the mirror?
Solution:
For a plane mirror, the deviation (\(\delta\)) is the angle between the incident ray and the reflected ray. The formula is:
\[\delta = 180^\circ - 2i\]
where \(i\) is the angle of incidence.
Given: \(i = 30^\circ\)
Substitute into the formula:
\[\delta = 180^\circ - 2 \times 30^\circ\]
\[\delta = 180^\circ - 60^\circ\]
\[\delta = 120^\circ\]
So, the deviation is 120°.
Answer: (1) 120°
2. The sides of a square sheet are increasing at the rate of 4 cm per second. Find the rate at which the area of the sheet is increasing when the side length is 8 cm.
Solution:
The area of a square is \(A = s^2\), where \(s\) is the side length.
We need to find \(\frac{dA}{dt}\), the rate of change of area with respect to time.
Differentiate \(A = s^2\) with respect to time \(t\):
\[\frac{dA}{dt} = 2s \cdot \frac{ds}{dt}\]
Given:
- \(\frac{ds}{dt} = 4 \, \text{cm/s}\) (rate of increase of side length)
- \(s = 8 \, \text{cm}\) (side length at that instant)
Substitute the values:
\[\frac{dA}{dt} = 2 \times 8 \times 4\]
\[\frac{dA}{dt} = 16 \times 4\]
\[\frac{dA}{dt} = 64 \, \text{cm}^2/\text{s}\]
So, the area is increasing at 64 cm²/s.
Answer: (4) 64 cm²/s
3. There is a convex mirror of radius 50 cm. The image of a point at a distance 50 cm from the pole of the mirror on its axis will be formed at:
Solution:
For a convex mirror:
- Focal length \(f = \frac{R}{2}\), where \(R\) is the radius of curvature, and \(f\) is positive.
Given: \(R = 50 \, \text{cm}\)
\[f = \frac{50}{2} = 25 \, \text{cm}\]
- Object distance \(u = -50 \, \text{cm}\) (negative, as the object is in front of the mirror).
Use the mirror formula:
\[\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\]
Substitute:
\[\frac{1}{25} = \frac{1}{-50} + \frac{1}{v}\]
\[\frac{1}{25} = -\frac{1}{50} + \frac{1}{v}\]
\[\frac{1}{v} = \frac{1}{25} + \frac{1}{50}\]
LCM of 25 and 50 is 50:
\[\frac{1}{v} = \frac{2}{50} + \frac{1}{50} = \frac{3}{50}\]
\[v = \frac{50}{3} \approx 16.67 \, \text{cm}\]
Since \(v\) is positive, the image is behind the mirror, approximately 16 cm.
Answer: (4) 16 cm behind the mirror
4. Two plane mirrors are at 45° to each other. If an object is placed between them, then the number of images will be:
Solution:
For two plane mirrors at an angle \(\theta\), the number of images \(n\) is given by:
\[n = \frac{360^\circ}{\theta} - 1\] if \(\frac{360^\circ}{\theta}\) is an even integer.
Given: \(\theta = 45^\circ\)
\[\frac{360^\circ}{45^\circ} = 8\]
Since 8 is even:
\[n = 8 - 1 = 7\]
So, the number of images is 7.
Answer: (3) 7
5. A convex mirror is used to form the image of a real object. Then which of the following statements is wrong?
Solution:
For a convex mirror with a real object:
- The image is always virtual (behind the mirror).
- The image is erect.
- The image is diminished (smaller than the object).
- The image lies between the pole and the focus (behind the mirror).
Check each option:
(1) True: The image is between the pole and focus.
(2) True: The image is diminished.
(3) True: The image is erect.
(4) False: The image is virtual, not real.
So, the wrong statement is (4).
Answer: (4) The image is real
6. A small candle 2.5 cm in size is placed 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed to receive a sharp image? What is the size of the image?
Solution:
For a concave mirror:
- Radius of curvature \(R = 36 \, \text{cm}\)
- Focal length \(f = -\frac{R}{2} = -\frac{36}{2} = -18 \, \text{cm}\) (negative for concave mirror).
- Object distance \(u = -27 \, \text{cm}\) (negative, as object is in front).
- Object height \(h = 2.5 \, \text{cm}\)
Use the mirror formula:
\[\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\]
\[\frac{1}{-18} = \frac{1}{-27} + \frac{1}{v}\]
\[-\frac{1}{18} = -\frac{1}{27} + \frac{1}{v}\]
\[\frac{1}{v} = -\frac{1}{18} + \frac{1}{27}\]
LCM of 18 and 27 is 54:
\[\frac{1}{v} = -\frac{3}{54} + \frac{2}{54} = -\frac{1}{54}\]
\[v = -54 \, \text{cm}\]
Since \(v\) is negative, the image is in front of the mirror (real), so the screen should be placed 54 cm from the mirror.
Magnification \(m = -\frac{v}{u}\):
\[m = -\frac{-54}{-27} = -2\]
Image height \(h' = m \times h\):
\[h' = -2 \times 2.5 = -5 \, \text{cm}\]
The image is inverted, and its size (magnitude) is 5 cm.
Answer: (1) 54, 5
7. The focal length of a concave mirror is 50 cm. Where should an object be placed so that its image is two times magnified, real, and inverted?
Solution:
For a concave mirror:
- Focal length \(f = -50 \, \text{cm}\) (negative for concave).
- Magnification \(m = -2\) (negative since the image is real and inverted).
\[m = -\frac{v}{u}\]
\[-2 = -\frac{v}{u}\]
\[\frac{v}{u} = 2\]
\[v = 2u\]
Since \(u\) is negative (object in front), and the image is real (\(v\) negative):
Use the mirror formula:
\[\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\]
\[\frac{1}{-50} = \frac{1}{u} + \frac{1}{2u}\]
\[-\frac{1}{50} = \frac{2 + 1}{2u} = \frac{3}{2u}\]
\[\frac{3}{2u} = -\frac{1}{50}\]
Cross-multiply:
\[3 \times 50 = -2u\]
\[150 = -2u\]
\[u = -75 \, \text{cm}\]
So, the object should be placed 75 cm in front of the mirror.
Answer: (1) 75 cm
8. A spherical mirror forms a real image of a point object placed in front. The distance of the image and object from the mirror is 30 cm and 0.2 m respectively. The focal length and nature of the mirror are:
Solution:
- Object distance \(u = -0.2 \, \text{m} = -20 \, \text{cm}\) (negative, in front).
- Image distance \(v = -30 \, \text{cm}\) (negative, real image in front).
Use the mirror formula:
\[\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\]
\[\frac{1}{f} = \frac{1}{-20} + \frac{1}{-30}\]
\[\frac{1}{f} = -\frac{1}{20} - \frac{1}{30}\]
LCM of 20 and 30 is 60:
\[\frac{1}{f} = -\frac{3}{60} - \frac{2}{60} = -\frac{5}{60} = -\frac{1}{12}\]
\[f = -12 \, \text{cm} = -120 \, \text{mm}\]
Since \(f\) is negative, the mirror is concave.
Focal length = 120 mm (magnitude), nature = concave.
Answer: (1) 120 mm; concave
9. A point object is placed at a distance of 10 cm and its real image is formed at a distance of 20 cm from a concave mirror. If the object is moved by 0.1 cm towards the mirror, the image will shift by about:
Solution:
Initial:
- \(u = -10 \, \text{cm}\), \(v = -20 \, \text{cm}\) (real image, in front).
\[\frac{1}{f} = \frac{1}{-10} + \frac{1}{-20} = -\frac{1}{10} - \frac{1}{20} = -\frac{3}{20}\]
\[f = -\frac{20}{3} \, \text{cm}\]
Use differentiation for small changes:
\[\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\]
Differentiate: \(-\frac{1}{v^2} \frac{dv}{du} - \frac{1}{u^2} = 0\)
\[\frac{dv}{du} = -\frac{v^2}{u^2}\]
\[\frac{v}{u} = \frac{-20}{-10} = 2\]
\[\frac{dv}{du} = -(2)^2 = -4\]
Object moves 0.1 cm towards mirror: \(u\) from -10 to -9.9 cm, so \(du = -9.9 - (-10) = 0.1 \, \text{cm}\).
\[dv = -4 \times 0.1 = -0.4 \, \text{cm}\]
Initial \(v = -20 \, \text{cm}\), new \(v = -20 - 0.4 = -20.4 \, \text{cm}\).
Image distance increases from 20 cm to 20.4 cm, so it shifts 0.4 cm away from the mirror.
Answer: (1) 0.4 cm away from the mirror
10. A convex lens forms a real image on a screen placed at a distance 60 cm from the object. When the lens is shifted towards the screen by 20 cm, another image of the object is formed on the screen. The focal length of the lens is:
Solution:
Object at \(x = 0\), image at \(x = 60 \, \text{cm}\).
Lens at \(x = d\):
- \(u = -d\)
- \(v = 60 - d\)
\[\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{60 - d} - \frac{1}{-d} = \frac{1}{60 - d} + \frac{1}{d}\]
\[f = \frac{d (60 - d)}{60}\]
Lens shifted to \(d + 20\):
- \(u' = -(d + 20)\)
-\(v' = 60 - (d + 20) = 40 - d\)
\[\frac{1}{f} = \frac{1}{40 - d} + \frac{1}{d + 20}\]
\[f = \frac{(d + 20)(40 - d)}{60}\]
Equate:
\[d (60 - d) = (d + 20)(40 - d)\]
\[60d - d^2 = 40d - d^2 + 800 - 20d\]
\[60d = 20d + 800\]
\[40d = 800\]
\[d = 20 \, \text{cm}\]
\[f = \frac{20 \times (60 - 20)}{60} = \frac{20 \times 40}{60} = \frac{800}{60} = \frac{40}{3} \, \text{cm}\]
Answer: (2) 40/3 cm
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