Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Geometrical Optics question

2019 · 12 Jan · Shift 2 · Q69
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Geometrical Optics
  5. /2019 · 12 Jan · Shift 2 · Q69

Geometrical Optics question

2019 · 12 Jan · Shift 2 · Q69

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1
Formation of real image using a biconvex lens is shown below : JEE Main 2019 (Online) 12th January Evening Slot Physics - Geometrical Optics Question 188 English If the whole set up is immersed in water without disturbing the object and the screen positions, what will one observe on the screen ?
  1. A
    Image disappears
  2. B
    Magnified image
  3. C
    Erect real image
  4. D
    No change
View written solutionFree

Correct answer: A

  1. What changes when the setup is immersed in water?

For a lens in a medium, the lens-maker formula is

1f=(nlensnmedium−1)(1R1−1R2)\frac{1}{f}=\left(\frac{n_{\text{lens}}}{n_{\text{medium}}}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)f1​=(nmedium​nlens​​−1)(R1​1​−R2​1​)

Initially, the lens is in air:

1fair=(nlensnair−1)(1R1−1R2)\frac{1}{f_{\text{air}}}=\left(\frac{n_{\text{lens}}}{n_{\text{air}}}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)fair​1​=(nair​nlens​​−1)(R1​1​−R2​1​)

After immersing in water:

1fwater=(nlensnwater−1)(1R1−1R2)\frac{1}{f_{\text{water}}}=\left(\frac{n_{\text{lens}}}{n_{\text{water}}}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)fwater​1​=(nwater​nlens​​−1)(R1​1​−R2​1​)

Since nwater>nairn_{\text{water}} > n_{\text{air}}nwater​>nair​, the factor

(nlensnwater−1)\left(\frac{n_{\text{lens}}}{n_{\text{water}}}-1\right)(nwater​nlens​​−1)

is smaller than

(nlensnair−1)\left(\frac{n_{\text{lens}}}{n_{\text{air}}}-1\right)(nair​nlens​​−1)

Hence the power of the lens decreases and its focal length increases:

fwater>fairf_{\text{water}} > f_{\text{air}}fwater​>fair​

So the biconvex lens becomes less converging in water.

  1. Effect on image position

The object and screen are kept fixed. Originally, in air, the screen was at the real image position, so a sharp real image was formed.

After immersion, because the focal length increases, the lens formula

1f=1v−1u\frac{1}{f}=\frac{1}{v}-\frac{1}{u}f1​=v1​−u1​

shows that for the same object distance uuu, the new image distance vvv changes.

Since the lens is now weaker, the real image shifts farther away from the lens (or may even fail to form on that side if the object lies within the new focal length). In either case, the screen kept at the old position will no longer coincide with the image position.

Therefore, a sharp image will not be seen on the screen.

  1. What is observed on the screen?

Because the image position changes while the screen remains fixed, the image is not formed on the screen. So effectively, the image disappears from the screen.

Thus, the correct option is:

A: Image disappears\boxed{\text{A: Image disappears}}A: Image disappears​
  1. Checking options
  • A: Image disappears — Correct.
  • B: Magnified image — Incorrect; screen is fixed, so a focused magnified image is not obtained there.
  • C: Erect real image — Incorrect; real images formed by a convex lens are inverted, not erect.
  • D: No change — Incorrect; refractive index of surrounding medium changes lens power.
  1. Comparison with stored answer

Stored correct answer: A

My derived answer: A

So they agree.

PreviousNext

More from Geometrical Optics

  • A particle is oscillating on the X-axis with an amplitude 2cm about the point x0​=10cm, with a frequency ω. A concave mirror of local length 5cm is placed at the origin (see figure). Identify the correct statements. Includes diagram2018 · MCQ
  • A planoconvex lens becomes an optical system of 28cm focal length when its plane surface is silvered and illuminated from left to right as shown in Fig-A. If the same lens is instead silvered on the curved surface and illuminated from… Includes diagram2018 · MCQ
  • A convergent doublet of separated lenses, corrected for spherical aberration, has resultant focal length of 10 cm. The separation between the two lenses is 2 cm. The focal lengths of the component lenses are :2018 · MCQ
  • A ray of light is incident at an angle of 60o on one face of a prism of angle 30o. The emergent ray of light makes an angle of 30o with incident ray. The angle made by the emergent ray with second face of prism will be :2018 · MCQ
  • Let the refractive index of a denser medium with respect to a rarer medium be n12 and its critical angle be θC . At an angle of incidence A when light is travelling from denser medium to rarer medium, a part of the light is reflected and…2017 · MCQ
  • In an experiment a convex lens of focal length 15 cm is placed coaxially on an optical bench in front of a convex mirror at a distance of 5 cm from it. It is found that an object and its image coincide, if the object is placed at a…2017 · MCQ
  • A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15cm from a converging lens of magnitude of focal length 20cm. A beam of parallel light falls on the diverging lens. The final image formed is:2017 · MCQ
  • To find the focal length of a convex mirror, a student records the following data : The focal length of the convex lens is f1 and that of mirror is f2. Then taking index correction to be negligibly small, f1 and f2 are close to : Includes table2016 · MCQ