Home
Class 12
PHYSICS
Assertion : The focal length of an equ...

Assertion : The focal length of an equiconvex lens placed in air is equal to radius of curvature of either face.
Reason : For an equiconvex lens radius of curvature of both the faces is same.

A

If both assertion and reason are true and reason is the correct explanation of assertion.

B

If both assertion and reason are true and reason is not the correct explanation of assertion.

C

If assertion is true but reason is false.

D

If both assertion and reason are false.

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the assertion and the reason provided in the question. ### Step 1: Understand the Assertion The assertion states that the focal length (f) of an equiconvex lens placed in air is equal to the radius of curvature (R) of either face of the lens. ### Step 2: Understand the Reason The reason states that for an equiconvex lens, the radius of curvature of both faces is the same. This is indeed true, as an equiconvex lens has two convex surfaces with equal radii of curvature. ### Step 3: Use the Lensmaker's Formula The lensmaker's formula is given by: \[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] where: - \( \mu \) is the refractive index of the lens material, - \( R_1 \) is the radius of curvature of the first surface, - \( R_2 \) is the radius of curvature of the second surface. For an equiconvex lens: - Let \( R_1 = R \) (for the first surface), - Let \( R_2 = -R \) (for the second surface, as per the sign convention). ### Step 4: Substitute Values into the Formula Substituting these values into the lensmaker's formula: \[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R} - \left(-\frac{1}{R}\right) \right) \] \[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R} + \frac{1}{R} \right) \] \[ \frac{1}{f} = (\mu - 1) \left( \frac{2}{R} \right) \] ### Step 5: Substitute the Refractive Index Assuming the refractive index \( \mu = 1.5 \): \[ \frac{1}{f} = (1.5 - 1) \left( \frac{2}{R} \right) \] \[ \frac{1}{f} = 0.5 \left( \frac{2}{R} \right) \] \[ \frac{1}{f} = \frac{1}{R} \] ### Step 6: Solve for Focal Length Taking the reciprocal gives: \[ f = R \] This confirms that the focal length of the equiconvex lens is indeed equal to the radius of curvature of either face. ### Conclusion Thus, both the assertion and the reason are true, but the reason does not correctly explain the assertion. The correct explanation is derived from the lensmaker's formula. ### Final Answer - **Assertion**: True - **Reason**: True, but not the correct explanation for the assertion. ---

To solve the problem, we need to analyze the assertion and the reason provided in the question. ### Step 1: Understand the Assertion The assertion states that the focal length (f) of an equiconvex lens placed in air is equal to the radius of curvature (R) of either face of the lens. ### Step 2: Understand the Reason The reason states that for an equiconvex lens, the radius of curvature of both faces is the same. This is indeed true, as an equiconvex lens has two convex surfaces with equal radii of curvature. ...
Promotional Banner

Topper's Solved these Questions

  • RAY OPTICS AND OPTICAL INSTRUMENTS

    NCERT FINGERTIPS ENGLISH|Exercise MULTIPLE CHOICE QUESTIONS|100 Videos
  • RAY OPTICS AND OPTICAL INSTRUMENTS

    NCERT FINGERTIPS ENGLISH|Exercise Reflection Of Light By Spherical Mirrors|5 Videos
  • RAY OPTICS AND OPTICAL INSTRUMENTS

    NCERT FINGERTIPS ENGLISH|Exercise NCERT EXAMPLAR PROBLEMS|11 Videos
  • PRACTICE PAPPER

    NCERT FINGERTIPS ENGLISH|Exercise Practice Paper 3|50 Videos
  • SEMICONDUCTOR ELECTRONICS : MATERIALS , DEVICES AND SIMPLE CIRCUITS

    NCERT FINGERTIPS ENGLISH|Exercise Assertion And Reason|15 Videos

Similar Questions

Explore conceptually related problems

The focal length of equiconvex lens of retractive index, mu=1.5 and radius of curvature, R is

The focal length of a concave mirror is 10 cm. Find its radius of curvature.

Find the focal length of the lens shown in Fig . The radii of curvature of both the surfaces are equal to R.

The focal length of a convex mirror is 20 cm its radius of curvature will be

A biconvex lens has focal length (2)/(3) times the radius of curvature of either surface. Calculate refraction index f material of the lens.

A biconvex lens has focal length (2)/(3) times the radius of curvature of either surface. Calculate refraction index f material of the lens.

The adjacent figure shows a thin plano-convex lens of refractive index mu_1 and a thin plano-concave lens of refractive index mu_2 , both having same radius of curvature R of their curved surfaces. The thin lens of refractive index mu_3 has radius of curvature R of both its surfaces. This lens is so placed in between the plano-convex and plano-concave lenses that the plane surfaces are parallel to each other. The focal length of the combination is

(b) Write the Lens Maker's formula and use it to obtain the range of mu (The refractive index of the material of the lens) for which the focal length of and equiconvex lens, kept in air, would have a greater magnitude than that of the radius of curvature of its two surfaces.

The focal length of an equi-concave lens is 3/4 times of radius of curvature of its surfaces. Find the refractive index of the material of lens. Under what condition will this lens behave as a converging lens?

Two converging glass lenses A and B have focal lengths in the ratio 2:1 . The radius of curvature of first surface of lens A is 1/4 th of the second surface where as the radius of curvature of first surface of lens B is twice that of second surface. Then the ratio between the radii of the first surfaces of A and B is